Chapter 9 Differential Equations

Given:

The family of curves is given by the equation:

where A and B are arbitrary constants.

Solution:

To find the differential equation, we need to eliminate the arbitrary constants A and B. Since there are two arbitrary constants, we need to differentiate the given equation twice with respect to x.

Differentiating equation (1) with respect to x, we get:

Using the derivative rule $\frac{d}{dx}(e^{ax}) = ae^{ax}$, we differentiate the terms:

Now, differentiating equation (3) with respect to x again, we get:

Applying the derivative rule again:

We can factor out 4 from the expression:

From equation (1), we have the original equation $y = Ae^{2x} + Be^{-2x}$. Substitute this expression for $(Ae^{2x} + Be^{-2x})$ into equation (5):

Rearranging the terms, we obtain the required differential equation:

Final Answer:

The differential equation for the given family of curves y = Ae$^{2x}$ + Be$^{-2x}$ is $\frac{d^2y}{dx^2}$ - 4y = 0.

Given:

The differential equation is $\frac{dy}{dx} = \frac{y}{x}$.

To Find:

The general solution of the given differential equation.

Solution:

The given differential equation is:

$\frac{dy}{dx} = \frac{y}{x}$

This is a first-order differential equation.

We can solve this by the method of separation of variables.

Rearrange the equation to separate the terms involving $y$ from the terms involving $x$:

$\frac{dy}{y} = \frac{dx}{x}$

Now, integrate both sides of the equation:

$\int \frac{1}{y} dy = \int \frac{1}{x} dx$

The integrals are standard forms:

$\int \frac{1}{y} dy = \log |y| + C_1$

$\int \frac{1}{x} dx = \log |x| + C_2$

So, integrating both sides gives:

$\log |y| = \log |x| + C$

where $C$ is the constant of integration ($C = C_2 - C_1$).

We can write the constant $C$ as $\log |A|$, where $A$ is an arbitrary positive constant.

$\log |y| = \log |x| + \log |A|$

Using the logarithm property $\log a + \log b = \log (ab)$, we get:

$\log |y| = \log |Ax|$

Taking the exponential of both sides:

$|y| = |Ax|$

This implies $y = \pm Ax$. Let $K = \pm A$. Since $A$ is an arbitrary positive constant, $K$ is an arbitrary non-zero constant.

If $y=0$, the original equation is satisfied ($\frac{d(0)}{dx} = 0$, $\frac{0}{x} = 0$). The solution $y=Kx$ includes $y=0$ if we allow $K=0$. Therefore, the general solution can be written as:

$y = Kx$

where $K$ is an arbitrary constant.

The general solution of the differential equation $\frac{dy}{dx} = \frac{y}{x}$ is $y = Kx$.

Given:

The differential equation is $\frac{dy}{dx} = ye^x$.

Initial condition: $x = 0$, $y = e$.

To Find:

The value of $y$ when $x = 1$.

Solution:

The given differential equation is:

$\frac{dy}{dx} = ye^x$

This is a first-order differential equation which can be solved by separation of variables. Assume $y \neq 0$ for separation. If $y=0$, then $\frac{dy}{dx}=0$, and $ye^x=0$, so $y=0$ is a trivial solution, but it does not satisfy the initial condition $y=e$ when $x=0$.

Separate the variables $y$ and $x$:

$\frac{dy}{y} = e^x dx$

Integrate both sides of the equation:

$\int \frac{1}{y} dy = \int e^x dx$

Performing the integration:

$\log |y| = e^x + C$

where $C$ is the constant of integration.

Now, we use the given initial condition $x = 0$, $y = e$ to find the value of $C$.

Substitute $x=0$ and $y=e$ into the general solution:

$\log |e| = e^0 + C$

$\log e = 1 + C$

Since $\log e = 1$ (assuming the natural logarithm, as $e^x$ suggests base $e$):

$1 = 1 + C$

Solving for $C$:

$C = 1 - 1$

$C = 0$

Substitute the value of $C$ back into the general solution to obtain the particular solution:

$\log |y| = e^x + 0$

$\log |y| = e^x$

To find $y$, take the exponential of both sides:

$|y| = e^{e^x}$

Since the initial condition has $y=e > 0$, the solution for $y$ will remain positive for values of $x$ where $e^{e^x}$ is defined. Thus, $|y|=y$.

$y = e^{e^x}$

Finally, we need to find the value of $y$ when $x = 1$. Substitute $x=1$ into the particular solution:

$y = e^{e^1}$

$y = e^e$

The value of $y$ when $x = 1$ is $e^e$.

Given:

The differential equation is $\frac{dy}{dx} + \frac{y}{x} = x^2$.

To Find:

The general solution of the given differential equation.

Solution:

The given differential equation is:

$\frac{dy}{dx} + \frac{y}{x} = x^2$

This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.

Comparing the given equation with the standard form, we have:

$P(x) = \frac{1}{x}$

$Q(x) = x^2$

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

First, calculate the integral of $P(x)$:

$\int P(x) dx = \int \frac{1}{x} dx = \log |x|$

The integrating factor is:

$IF = e^{\int P(x) dx} = e^{\log |x|} = |x|$

Assuming $x > 0$, we can take $IF = x$.

Multiply the given differential equation by the integrating factor, $x$:

$x \left(\frac{dy}{dx} + \frac{y}{x}\right) = x (x^2)$

$x \frac{dy}{dx} + x \cdot \frac{y}{x} = x^3$

$x \frac{dy}{dx} + y = x^3$

The left side of the equation is the derivative of the product $(xy)$ with respect to $x$, i.e., $\frac{d}{dx}(xy) = x \frac{dy}{dx} + y$.

So, the equation becomes:

$\frac{d}{dx}(xy) = x^3$

Now, integrate both sides with respect to $x$:

$\int \frac{d}{dx}(xy) dx = \int x^3 dx$

$xy = \frac{x^{3+1}}{3+1} + C$

$xy = \frac{x^4}{4} + C$

where $C$ is the constant of integration.

Finally, solve for $y$ by dividing both sides by $x$ (for $x \neq 0$):

$y = \frac{1}{x} \left(\frac{x^4}{4} + C\right)$

$y = \frac{x^4}{4x} + \frac{C}{x}$

$y = \frac{x^3}{4} + \frac{C}{x}$

The general solution of the differential equation is $y = \frac{x^3}{4} + \frac{C}{x}$, where $C$ is an arbitrary constant.

Given:

A family of lines passing through the origin.

To Find:

The differential equation of this family of lines.

Solution:

The general equation of a line is $y = mx + c$.

Since the line passes through the origin $(0,0)$, the point $(0,0)$ must satisfy the equation of the line.

Substitute $x=0$ and $y=0$ into the general equation:

$0 = m(0) + c$

$0 = 0 + c$

$c = 0$

Thus, the equation of any line passing through the origin is given by:

where $m$ is the slope and represents the arbitrary constant of the family.

Since there is only one arbitrary constant ($m$), we need to differentiate the equation (i) once with respect to $x$ to eliminate $m$.

Differentiating equation (i) with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(mx)$

Now, substitute the value of $m$ from equation (ii) into equation (i) to eliminate the arbitrary constant $m$.

Substitute $m = \frac{dy}{dx}$ into $y = mx$:

$y = \left(\frac{dy}{dx}\right) x$

Rearrange the equation to get the differential equation:

$y = x \frac{dy}{dx}$

or

$x \frac{dy}{dx} - y = 0$

This is the differential equation of the family of lines passing through the origin.

Given:

A family of all non-horizontal lines in a plane.

To Find:

The differential equation of this family of lines.

Solution:

The general equation of a line in a plane is given by:

$Ax + By + C = 0$

We consider lines that can be expressed in the form $y = mx + c$. This form represents all lines except vertical lines (where the slope is undefined).

Horizontal lines have a slope $m = 0$. Non-horizontal lines, represented in this form, must have a slope $m \neq 0$. Thus, the family of non-horizontal lines (excluding vertical lines) is given by:

where $m$ and $c$ are arbitrary constants, and the condition $m \neq 0$ must be satisfied.

Since there are two arbitrary constants ($m$ and $c$) in equation (i), we need to differentiate twice with respect to $x$ to eliminate them.

Differentiate equation (i) with respect to $x$ once:

$\frac{dy}{dx} = \frac{d}{dx}(mx + c)$

Differentiate equation (ii) with respect to $x$ again:

$\frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(m)$

$\frac{d^2y}{dx^2} = 0$

This differential equation, $\frac{d^2y}{dx^2} = 0$, is the differential equation for all lines of the form $y = mx + c$, including horizontal lines (where $m=0$).

We are asked for the differential equation of all non-horizontal lines. This means the slope must not be zero. From equation (ii), the slope is $m = \frac{dy}{dx}$.

So, for non-horizontal lines, we must have $m \neq 0$, which means:

$\frac{dy}{dx} \neq 0$

Therefore, the differential equation for all non-horizontal lines that can be represented as $y=f(x)$ is $\frac{d^2y}{dx^2} = 0$ with the condition $\frac{dy}{dx} \neq 0$.

Note: This differential equation does not include vertical lines (e.g., $x = k$), which are also non-horizontal. The differential equation for a vertical line $x=k$ is $\frac{dx}{dy} = 0$. However, when asking for the differential equation in terms of $\frac{dy}{dx}, \frac{d^2y}{dx^2}$, etc., the context usually implies curves where $y$ is a function of $x$.

Given:

The slope of the tangent at any point $(x, y)$ on the curve, different from the origin, is given by $y + \frac{y}{x}$.

To Find:

The equation of the curve.

Solution:

The slope of the tangent to a curve at any point $(x, y)$ is given by the derivative $\frac{dy}{dx}$.

According to the problem statement, the slope is $y + \frac{y}{x}$ for $(x,y) \neq (0,0)$.

So, the differential equation of the curve is:

We can rewrite the equation (i) by factoring $y$ on the right side:

$\frac{dy}{dx} = y \left(1 + \frac{1}{x}\right)$

This is a first-order differential equation that can be solved by separating the variables. Assume $y \neq 0$ and $x \neq 0$. If $y=0$, the equation becomes $\frac{dy}{dx}=0$, which gives $y=C$. Substituting into the original equation gives $0 = 0 + 0$, so $y=0$ is a solution. This corresponds to the x-axis, which passes through the origin, but its tangent slope at any point $(x,0)$ with $x \neq 0$ is $0 + \frac{0}{x} = 0$, which matches the derivative $\frac{d(0)}{dx}=0$. So $y=0$ is a valid curve.

For $y \neq 0$ and $x \neq 0$, separate the variables:

$\frac{dy}{y} = \left(1 + \frac{1}{x}\right) dx$

Now, integrate both sides of the equation:

$\int \frac{1}{y} dy = \int \left(1 + \frac{1}{x}\right) dx$

Performing the integration:

$\int \frac{1}{y} dy = \log |y| + C_1$

$\int \left(1 + \frac{1}{x}\right) dx = \int 1 dx + \int \frac{1}{x} dx = x + \log |x| + C_2$

Combining the results:

$\log |y| = x + \log |x| + C$

where $C = C_2 - C_1$ is the constant of integration.

Rearrange the terms to isolate the constant:

$\log |y| - \log |x| = x + C$

Using the logarithm property $\log a - \log b = \log (a/b)$, we get:

$\log \left|\frac{y}{x}\right| = x + C$

To solve for $\frac{y}{x}$, take the exponential of both sides:

$\left|\frac{y}{x}\right| = e^{x+C}$

$\left|\frac{y}{x}\right| = e^x \cdot e^C$

Let $e^C = A_1$, where $A_1$ is a positive constant. Then:

$\left|\frac{y}{x}\right| = A_1 e^x$

This means $\frac{y}{x} = \pm A_1 e^x$.

Let $A = \pm A_1$. $A$ is an arbitrary non-zero constant.

$\frac{y}{x} = A e^x$

Solve for $y$:

$y = Ax e^x$

As noted earlier, $y=0$ is also a solution. The equation $y = Axe^x$ includes the solution $y=0$ when $A=0$. Therefore, the general solution represents the family of curves, including the case $A=0$.

The condition "tangent at any point on it, different from origin" specifies the domain where the slope formula applies, which is consistent with our derivation for $x \neq 0$.

The equation of the curve is $y = Axe^x$, where $A$ is an arbitrary constant.

Given:

A curve passes through the point (1, 1).

The perpendicular distance of the origin from the normal at any point P(x, y) on the curve (for P different from origin) is equal to the distance of P from the x-axis.

To Find:

The equation of the curve.

Solution:

Let P(x, y) be any point on the curve, different from the origin (0, 0).

The slope of the tangent at the point P(x, y) is $\frac{dy}{dx}$.

The slope of the normal at the point P(x, y) is $m_{normal} = -\frac{1}{\frac{dy}{dx}}$, assuming $\frac{dy}{dx} \neq 0$. If $\frac{dy}{dx} = 0$, the tangent is horizontal and the normal is vertical, $x = \text{constant}$. If the tangent is vertical, $\frac{dy}{dx}$ is undefined, and the normal is horizontal, $y = \text{constant}$. We will proceed assuming the slope is well-defined and check boundary cases later.

The equation of the normal line at P(x, y) is given by:

$Y - y = m_{normal} (X - x)$

$Y - y = -\frac{1}{\frac{dy}{dx}} (X - x)$

Assuming $\frac{dy}{dx} \neq 0$, multiply both sides by $\frac{dy}{dx}$:

$(Y - y) \frac{dy}{dx} = -(X - x)$

$Y \frac{dy}{dx} - y \frac{dy}{dx} = -X + x$

Rearrange the equation into the form $AX + BY + C = 0$:

$X + Y \frac{dy}{dx} - x - y \frac{dy}{dx} = 0$

The perpendicular distance from the origin (0, 0) to the normal line $X + \frac{dy}{dx} Y + (-x - y \frac{dy}{dx}) = 0$ is given by the formula $\frac{|AX_0 + BY_0 + C|}{\sqrt{A^2 + B^2}}$ where $(X_0, Y_0) = (0, 0)$.

Distance from origin $= \frac{|1(0) + \frac{dy}{dx}(0) + (-x - y \frac{dy}{dx})|}{\sqrt{1^2 + (\frac{dy}{dx})^2}}$

Distance from origin $= \frac{|-x - y \frac{dy}{dx}|}{\sqrt{1 + (\frac{dy}{dx})^2}}$

Distance from origin $= \frac{|x + y \frac{dy}{dx}|}{\sqrt{1 + (\frac{dy}{dx})^2}}$

The distance of the point P(x, y) from the x-axis is $|y|$.

According to the problem statement, these two distances are equal:

Square both sides of equation (i):

$\frac{(x + y \frac{dy}{dx})^2}{1 + (\frac{dy}{dx})^2} = y^2$

$(x + y \frac{dy}{dx})^2 = y^2 (1 + (\frac{dy}{dx})^2)$

$x^2 + 2xy \frac{dy}{dx} + y^2 (\frac{dy}{dx})^2 = y^2 + y^2 (\frac{dy}{dx})^2$

Cancel the term $y^2 (\frac{dy}{dx})^2$ from both sides:

$x^2 + 2xy \frac{dy}{dx} = y^2$

Rearrange to get a differential equation:

$2xy \frac{dy}{dx} = y^2 - x^2$

$\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$

This is a homogeneous differential equation. We can solve it by substituting $y = vx$. Then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.

Substitute $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation:

$v + x \frac{dv}{dx} = \frac{(vx)^2 - x^2}{2x(vx)}$

$v + x \frac{dv}{dx} = \frac{v^2 x^2 - x^2}{2vx^2}$

$v + x \frac{dv}{dx} = \frac{x^2(v^2 - 1)}{2vx^2}$

$v + x \frac{dv}{dx} = \frac{v^2 - 1}{2v}$

Isolate $x \frac{dv}{dx}$:

$x \frac{dv}{dx} = \frac{v^2 - 1}{2v} - v$

$x \frac{dv}{dx} = \frac{v^2 - 1 - 2v^2}{2v}$

$x \frac{dv}{dx} = \frac{-v^2 - 1}{2v}$

$x \frac{dv}{dx} = -\frac{v^2 + 1}{2v}$

Separate the variables $v$ and $x$:

$\frac{2v}{v^2 + 1} dv = -\frac{1}{x} dx$

Integrate both sides:

$\int \frac{2v}{v^2 + 1} dv = \int -\frac{1}{x} dx$

$\log |v^2 + 1| = -\log |x| + C$

where $C$ is the constant of integration.

Since $v^2 + 1 > 0$, we can write $\log (v^2 + 1)$.

$\log (v^2 + 1) = -\log |x| + C$

$\log (v^2 + 1) + \log |x| = C$

Using logarithm properties:

$\log (|x|(v^2 + 1)) = C$

Exponentiate both sides:

$|x|(v^2 + 1) = e^C$

Let $e^C = A_1$, where $A_1$ is a positive constant.

$|x|(v^2 + 1) = A_1$

Substitute back $v = \frac{y}{x}$:

$|x|\left(\left(\frac{y}{x}\right)^2 + 1\right) = A_1$

$|x|\left(\frac{y^2}{x^2} + 1\right) = A_1$

$|x|\left(\frac{y^2 + x^2}{x^2}\right) = A_1$

$\frac{|x|}{x^2}(y^2 + x^2) = A_1$

$\frac{1}{|x|}(y^2 + x^2) = A_1$

$y^2 + x^2 = A_1 |x|$

Let $A$ be an arbitrary constant such that $A = A_1$ if $x>0$ and $A = -A_1$ if $x<0$. Note that $A \neq 0$ since $A_1 > 0$. However, the case $y=0$ led to $x^2 = A_1|x|$, which includes $x^2 = Ax$ or $x^2 = -Ax$. $x^2-Ax=0 \implies x(x-A)=0 \implies x=0$ or $x=A$. This represents lines $x=0$ or $x=A$ (if $y=0$). The question implies $y=f(x)$ type curve.

Let's keep the form $y^2 + x^2 = A_1 |x|$. Since the curve passes through (1, 1), where $x=1>0$, we use $|x|=x$ in the vicinity of (1,1).

The general form of the curve is $y^2 + x^2 = Ax$, where $A$ is an arbitrary constant.

$x^2 - Ax + y^2 = 0$

Now, use the given initial condition that the curve passes through the point (1, 1). Substitute $x=1$ and $y=1$ into the equation:

$1^2 - A(1) + 1^2 = 0$

$1 - A + 1 = 0$

$2 - A = 0$

$A = 2$

Substitute the value of $A$ back into the equation of the curve:

$x^2 - 2x + y^2 = 0$

This equation represents a circle. We can complete the square for the $x$ terms to find its center and radius:

$(x^2 - 2x + 1) + y^2 = 0 + 1$

$(x - 1)^2 + y^2 = 1^2$

This is the equation of a circle with center (1, 0) and radius 1.

The equation of the curve is $x^2 - 2x + y^2 = 0$. This curve passes through (1,1). Note that the origin (0,0) also lies on this circle, but the condition in the problem is applied to points different from the origin.

Given:

The slope of the tangent to the curve at any point P(x, y) is $\frac{y}{x} - \cos^2 \frac{y}{x}$.

The curve passes through the point $\left( 1, \frac{π}{4} \right)$.

To Find:

The equation of the curve.

Solution:

The slope of the tangent at any point (x, y) on the curve is given by $\frac{dy}{dx}$.

According to the problem, the differential equation is:

$\frac{dy}{dx} = \frac{y}{x} - \cos^2 \frac{y}{x}$

This is a homogeneous differential equation of the form $\frac{dy}{dx} = f\left(\frac{y}{x}\right)$.

We use the substitution $y = vx$, where $v$ is a function of $x$.

Differentiating $y = vx$ with respect to $x$, we get:

$\frac{dy}{dx} = v + x \frac{dv}{dx}$

Substitute $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation:

$v + x \frac{dv}{dx} = \frac{vx}{x} - \cos^2 \frac{vx}{x}$

$v + x \frac{dv}{dx} = v - \cos^2 v$

Subtract $v$ from both sides:

$x \frac{dv}{dx} = -\cos^2 v$

This is a separable differential equation. Assuming $\cos v \neq 0$, we can separate the variables:

$\frac{dv}{\cos^2 v} = -\frac{dx}{x}$

$\sec^2 v \, dv = -\frac{1}{x} dx$

Integrate both sides:

$\int \sec^2 v \, dv = \int -\frac{1}{x} dx$

$\tan v = -\log |x| + C$

where $C$ is the constant of integration.

Substitute back $v = \frac{y}{x}$:

This is the general solution of the differential equation.

We are given that the curve passes through the point $\left( 1, \frac{π}{4} \right)$. We use this initial condition to find the value of $C$.

Substitute $x = 1$ and $y = \frac{π}{4}$ into equation (i):

$\tan \left(\frac{π/4}{1}\right) = -\log |1| + C$

$\tan \left(\frac{π}{4}\right) = -\log 1 + C$

Since $\tan \left(\frac{π}{4}\right) = 1$ and $\log 1 = 0$:

$1 = -0 + C$

$C = 1$

Substitute the value of $C=1$ back into the general solution (i) to get the equation of the curve (the particular solution):

$\tan \left(\frac{y}{x}\right) = -\log |x| + 1$

Since the point $(1, \frac{π}{4})$ has $x=1>0$, we can replace $|x|$ with $x$ for the curve passing through this point in the region where $x>0$.

The equation of the curve is $\tan \left(\frac{y}{x}\right) = 1 - \log x$.

Given:

The differential equation is $x^2 \frac{dy}{dx} - xy = 1 + \cos \left( \frac{y}{x} \right)$, with $x \neq 0$.

The initial condition is $x = 1$, $y = \frac{π}{2}$.

To Find:

The particular solution (equation of the curve) satisfying the given initial condition.

Solution:

The given differential equation is:

$x^2 \frac{dy}{dx} - xy = 1 + \cos \left( \frac{y}{x} \right)$

Rearrange the equation to get $\frac{dy}{dx}$ on one side:

$x^2 \frac{dy}{dx} = xy + 1 + \cos \left( \frac{y}{x} \right)$

$\frac{dy}{dx} = \frac{xy + 1 + \cos \left( \frac{y}{x} \right)}{x^2}$

$\frac{dy}{dx} = \frac{xy}{x^2} + \frac{1}{x^2} + \frac{1}{x^2} \cos \left( \frac{y}{x} \right)$

$\frac{dy}{dx} = \frac{y}{x} + \frac{1}{x^2} \left(1 + \cos \left( \frac{y}{x} \right)\right)$

This equation is not directly homogeneous. Let's check the substitution $v = \frac{y}{x}$, so $y = vx$. Then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.

Substitute into the original equation $x^2 \frac{dy}{dx} - xy = 1 + \cos \left( \frac{y}{x} \right)$:

$x^2 \left(v + x \frac{dv}{dx}\right) - x(vx) = 1 + \cos(v)$

$x^2 v + x^3 \frac{dv}{dx} - x^2 v = 1 + \cos(v)$

$x^3 \frac{dv}{dx} = 1 + \cos(v)$

This is a separable differential equation. Assuming $1 + \cos(v) \neq 0$, separate the variables $v$ and $x$:

$\frac{dv}{1 + \cos(v)} = \frac{dx}{x^3}$

Integrate both sides:

$\int \frac{dv}{1 + \cos(v)} = \int x^{-3} dx$

For the left integral, use the identity $1 + \cos(v) = 2 \cos^2 \left(\frac{v}{2}\right)$:

$\int \frac{dv}{2 \cos^2 \left(\frac{v}{2}\right)} = \int \frac{1}{2} \sec^2 \left(\frac{v}{2}\right) dv$

Let $u = \frac{v}{2}$, so $du = \frac{1}{2} dv$. The integral becomes:

$\int \frac{1}{2} \sec^2(u) (2 du) = \int \sec^2(u) du = \tan(u) + C_1 = \tan\left(\frac{v}{2}\right) + C_1$

For the right integral:

$\int x^{-3} dx = \frac{x^{-2}}{-2} + C_2 = -\frac{1}{2x^2} + C_2$

Equating the results of the integrals:

$\tan\left(\frac{v}{2}\right) = -\frac{1}{2x^2} + C$

where $C = C_2 - C_1$ is the integration constant.

Substitute back $v = \frac{y}{x}$:

This is the general solution.

Now, we use the initial condition $x = 1, y = \frac{π}{2}$ to find the value of $C$. Substitute these values into equation (i):

$\tan\left(\frac{π/2}{2 \cdot 1}\right) = -\frac{1}{2(1)^2} + C$

$\tan\left(\frac{π}{4}\right) = -\frac{1}{2} + C$

Since $\tan\left(\frac{π}{4}\right) = 1$:

$1 = -\frac{1}{2} + C$

$C = 1 + \frac{1}{2} = \frac{3}{2}$

Substitute the value of $C = \frac{3}{2}$ back into the general solution (i) to obtain the particular solution:

$\tan\left(\frac{y}{2x}\right) = -\frac{1}{2x^2} + \frac{3}{2}$

$\tan\left(\frac{y}{2x}\right) = \frac{3}{2} - \frac{1}{2x^2}$

$\tan\left(\frac{y}{2x}\right) = \frac{3x^2 - 1}{2x^2}$

This is the equation of the curve passing through the point $\left( 1, \frac{π}{4} \right)$.

Note: The assumption $1+\cos(v) \neq 0$ corresponds to $\cos(v) \neq -1$, which means $v \neq (2n+1)π$ for integer $n$. Since $v = y/x$, this means $y/x \neq (2n+1)π$. For the given initial condition $x=1, y=π/2$, $y/x = π/2$, and $\cos(π/2) = 0$, so $1+\cos(π/2) = 1 \neq 0$. Thus the separation of variables step is valid at the given point.

Given:

The differential equation is $x\;dy \;–\; y\;dx = \sqrt{x^2+y^2} \;dx$, with $x ≠ 0$.

To Find:

The type and the general solution of the given differential equation.

Solution:

The given differential equation is:

$x\;dy \;–\; y\;dx = \sqrt{x^2+y^2} \;dx$

Rearrange the equation to express $\frac{dy}{dx}$:

$x\;dy = y\;dx + \sqrt{x^2+y^2} \;dx$

$x\;dy = (y + \sqrt{x^2+y^2}) \;dx$

Let $f(x, y) = \frac{y + \sqrt{x^2+y^2}}{x}$. We check if the function is homogeneous by evaluating $f(\lambda x, \lambda y)$ for $\lambda \neq 0$:

$f(\lambda x, \lambda y) = \frac{\lambda y + \sqrt{(\lambda x)^2+(\lambda y)^2}}{\lambda x} = \frac{\lambda y + \sqrt{\lambda^2 x^2+\lambda^2 y^2}}{\lambda x}$

$f(\lambda x, \lambda y) = \frac{\lambda y + \sqrt{\lambda^2(x^2+y^2)}}{\lambda x} = \frac{\lambda y + |\lambda|\sqrt{x^2+y^2}}{\lambda x}$

For $\lambda > 0$, $|\lambda| = \lambda$.

$f(\lambda x, \lambda y) = \frac{\lambda y + \lambda\sqrt{x^2+y^2}}{\lambda x} = \frac{\lambda (y + \sqrt{x^2+y^2})}{\lambda x} = \frac{y + \sqrt{x^2+y^2}}{x} = f(x, y)$

Since $f(\lambda x, \lambda y) = \lambda^0 f(x, y)$ (for $\lambda > 0$), the differential equation is a homogeneous differential equation.

To solve this homogeneous equation, we use the substitution $y = vx$, where $v$ is a function of $x$.

Differentiating with respect to $x$, we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.

Substitute $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation (i):

$v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^2+(vx)^2}}{x}$

$v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^2(1+v^2)}}{x}$

$v + x \frac{dv}{dx} = \frac{vx + |x|\sqrt{1+v^2}}{x}$

$v + x \frac{dv}{dx} = v + \frac{|x|}{x}\sqrt{1+v^2}$

$x \frac{dv}{dx} = \frac{|x|}{x}\sqrt{1+v^2}$

Separate the variables $v$ and $x$. Assume $\sqrt{1+v^2} \neq 0$, which is always true for real $v$.

$\frac{dv}{\sqrt{1+v^2}} = \frac{|x|}{x^2} dx$

Integrate both sides:

$\int \frac{dv}{\sqrt{1+v^2}} = \int \frac{|x|}{x^2} dx$

The left side integral is $\log|v + \sqrt{1+v^2}|$.

The right side integral $\int \frac{|x|}{x^2} dx$ can be evaluated. For $x>0$, it is $\int \frac{1}{x} dx = \log|x| + C_1$. For $x<0$, it is $\int -\frac{1}{x} dx = -\log|x| + C_2$. We can combine these results using a constant. Let the integration result be $\log|Kx|$ for some non-zero constant $K$ (absorbing the sign of $x$ and the constant of integration).

So, we have:

$\log|v + \sqrt{1+v^2}| = \log|Kx|$

Exponentiating both sides:

$|v + \sqrt{1+v^2}| = |Kx|$

$v + \sqrt{1+v^2} = \pm Kx$. Let $C = \pm K$, where $C$ is a non-zero constant.

$v + \sqrt{1+v^2} = Cx$

Substitute back $v = \frac{y}{x}$:

$\frac{y}{x} + \sqrt{1+\left(\frac{y}{x}\right)^2} = Cx$

$\frac{y}{x} + \sqrt{\frac{x^2+y^2}{x^2}} = Cx$

$\frac{y}{x} + \frac{\sqrt{x^2+y^2}}{|x|} = Cx$

Multiply the equation by $|x|$:

$\frac{|x|}{x} y + |x| \frac{\sqrt{x^2+y^2}}{|x|} = Cx|x|$

$\text{sgn}(x) y + \sqrt{x^2+y^2} = Cx|x|$

This gives two possibilities:

Case 1: If $x > 0$, then $\text{sgn}(x)=1$ and $|x|=x$. The equation becomes $y + \sqrt{x^2+y^2} = Cx^2$.

Case 2: If $x < 0$, then $\text{sgn}(x)=-1$ and $|x|=-x$. The equation becomes $-y + \sqrt{x^2+y^2} = Cx(-x) = -Cx^2$, which can be written as $y - \sqrt{x^2+y^2} = Cx^2$.

So the general solution is of the form $y \pm \sqrt{x^2+y^2} = Cx^2$ where $C \neq 0$ is an arbitrary constant (the sign depends on the part of the curve). Let's rearrange this form.

Consider $y \pm \sqrt{x^2+y^2} = Cx^2$. This is equivalent to $\pm \sqrt{x^2+y^2} = Cx^2 - y$.

Squaring both sides gives:

$x^2+y^2 = (Cx^2 - y)^2$

$x^2+y^2 = C^2 x^4 - 2Cx^2 y + y^2$

$x^2 = C^2 x^4 - 2Cx^2 y$

Since $x \neq 0$, we can divide by $x^2$:

$1 = C^2 x^2 - 2Cy$

Rearrange to solve for $y$:

$2Cy = C^2 x^2 - 1$

$y = \frac{C^2 x^2 - 1}{2C}$

$y = \frac{C^2}{2C} x^2 - \frac{1}{2C}$

$y = \frac{C}{2} x^2 - \frac{1}{2C}$

Let $A = \frac{C}{2}$. Since $C$ is an arbitrary non-zero constant, $A$ is also an arbitrary non-zero constant. Then $-\frac{1}{2C} = -\frac{1}{2(2A)} = -\frac{1}{4A}$.

Thus, the general solution can be written as:

where $A$ is an arbitrary non-zero constant.

Given:

The differential equation $\left( 1 + \frac{dy}{dx} \right)^3 = \left( \frac{d^2y}{dx^2} \right)^2$.

To Find:

The degree of the differential equation.

Solution:

The given differential equation is:

$\left( 1 + \frac{dy}{dx} \right)^3 = \left( \frac{d^2y}{dx^2} \right)^2$

The order of a differential equation is the order of the highest derivative appearing in the equation.

In the given equation, the highest order derivative is $\frac{d^2y}{dx^2}$, which is a second-order derivative.

Thus, the order of the differential equation is 2.

The degree of a differential equation is the power of the highest order derivative, after the equation has been made free from radicals and fractions as far as the derivatives are concerned.

In the given equation, $\left( 1 + \frac{dy}{dx} \right)^3 = \left( \frac{d^2y}{dx^2} \right)^2$, the derivatives appear with integer powers (3 and 2), and there are no radicals or fractions involving derivatives.

The highest order derivative is $\frac{d^2y}{dx^2}$. Its power in the equation is 2.

The power of the lower order derivative $\frac{dy}{dx}$ is 3, but this does not determine the degree.

Therefore, the degree of the differential equation is 2.

The correct option is (B).

Given:

The differential equation $\frac{d^2y}{dx^2} + 3 \left( \frac{dy}{dx} \right)^2 = x^2 \log \left( \frac{d^2y}{dx^2} \right)$.

To Find:

The degree of the differential equation.

Solution:

The given differential equation is:

$\frac{d^2y}{dx^2} + 3 \left( \frac{dy}{dx} \right)^2 = x^2 \log \left( \frac{d^2y}{dx^2} \right)$

The order of the differential equation is the order of the highest derivative present, which is $\frac{d^2y}{dx^2}$.

The order is 2.

The degree of a differential equation is defined as the power of the highest order derivative, provided the equation is a polynomial equation in derivatives.

In the given equation, the term $\log \left( \frac{d^2y}{dx^2} \right)$ involves the second derivative inside a transcendental function (logarithm).

This equation cannot be expressed as a polynomial in terms of its derivatives $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$.

When a differential equation cannot be written as a polynomial in its derivatives, its degree is not defined.

Therefore, the degree of the given differential equation is not defined.

The correct option is (D).

Given:

The differential equation $\left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^2 = \frac{d^2y}{dx^2}$.

To Find:

The order and degree of the differential equation.

Solution:

The given differential equation is:

$\left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^2 = \frac{d^2y}{dx^2}$

The order of the differential equation is the order of the highest derivative appearing in the equation.

The derivatives present are $\frac{dy}{dx}$ (first order) and $\frac{d^2y}{dx^2}$ (second order).

The highest order derivative is $\frac{d^2y}{dx^2}$.

Thus, the order of the differential equation is 2.

The degree of a differential equation is the power of the highest order derivative, after the equation has been made free from radicals and fractions as far as the derivatives are concerned.

The equation is $\left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^2 = \frac{d^2y}{dx^2}$.

The term involving the highest order derivative is $\frac{d^2y}{dx^2}$. Its power in the equation is 1.

The equation is already a polynomial in terms of the derivatives.

The term $\left( \frac{dy}{dx} \right)^2$ is raised to the power of 2, giving $(\frac{dy}{dx})^4$ if expanded, but this is a lower order derivative. The degree is determined by the power of the highest order derivative.

The power of the highest order derivative $\frac{d^2y}{dx^2}$ is 1.

Therefore, the degree of the differential equation is 1.

The order is 2 and the degree is 1.

The correct option is (C).

Given:

A family of circles with a given radius $a$.

To Find:

The order of the differential equation representing this family of circles.

Solution:

The general equation of a circle with center $(h, k)$ and radius $a$ is given by:

$(x - h)^2 + (y - k)^2 = a^2$

In this equation, $x$ and $y$ are variables, and $a$ is a given constant (not an arbitrary parameter).

The arbitrary constants are the coordinates of the center, $h$ and $k$.

The number of independent arbitrary constants in the equation of the family of curves is equal to the order of the corresponding differential equation.

In the equation $(x - h)^2 + (y - k)^2 = a^2$, there are two independent arbitrary constants, $h$ and $k$.

To find the differential equation, we would differentiate the equation twice with respect to $x$ to eliminate the two arbitrary constants $h$ and $k$.

Differentiating once:

$2(x-h) + 2(y-k)\frac{dy}{dx} = 0$

$(x-h) + (y-k)\frac{dy}{dx} = 0$

Differentiating a second time:

$1 + (y-k)\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)\left(\frac{dy}{dx}\right) = 0$

$1 + (y-k)\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0$

From the first differentiated equation, $(x-h) = -(y-k)\frac{dy}{dx}$. We would then need to eliminate $h$ and $k$ using these two differentiated equations along with the original equation. The highest order derivative involved in this process will be $\frac{d^2y}{dx^2}$.

Since there are two arbitrary constants ($h$ and $k$), the order of the differential equation is 2.

The correct option is (B).

Given:

The differential equation $2x \frac{dy}{dx} − y = 3$.

To Find:

The type of curve represented by the general solution of the differential equation.

Solution:

The given differential equation is:

$2x \frac{dy}{dx} − y = 3$

Rearrange the equation to the standard form of a first-order linear differential equation, $\frac{dy}{dx} + P(x)y = Q(x)$.

$2x \frac{dy}{dx} = y + 3$

$\frac{dy}{dx} = \frac{y + 3}{2x}$

This is also a separable differential equation. We can separate the variables by moving terms involving $y$ to one side and terms involving $x$ to the other. Assume $y+3 \neq 0$. If $y+3=0$, then $y=-3$, and $\frac{dy}{dx}=0$. Substituting into the original equation gives $2x(0) - (-3) = 3$, which is $3=3$. So $y=-3$ is a solution, which is a horizontal line.

For $y+3 \neq 0$:

$\frac{dy}{y + 3} = \frac{dx}{2x}$

Integrate both sides:

$\int \frac{1}{y + 3} dy = \int \frac{1}{2x} dx$

$\log |y + 3| = \frac{1}{2} \log |x| + C'$

where $C'$ is the constant of integration.

Rewrite the constant $C'$ as $\frac{1}{2} \log |C|$, where $C$ is an arbitrary positive constant. This form helps simplify the solution.

$\log |y + 3| = \frac{1}{2} \log |x| + \frac{1}{2} \log |C|$

$\log |y + 3| = \frac{1}{2} (\log |x| + \log |C|)$

$\log |y + 3| = \frac{1}{2} \log |Cx|$

$\log |y + 3| = \log \sqrt{|Cx|}$

Exponentiate both sides:

$|y + 3| = \sqrt{|Cx|}$

Squaring both sides (this might introduce extraneous solutions, but we'll see):

$(y + 3)^2 = |Cx|$

$(y + 3)^2 = Kx$, where $K = \pm C$, an arbitrary non-zero constant.

The equation is $(y + 3)^2 = Kx$. This is the standard form of a parabola with its axis parallel to the x-axis and vertex at $(0, -3)$.

We should also consider the case $y=-3$. As checked earlier, $y=-3$ is a solution. This is a horizontal line. However, the general solution $(y+3)^2 = Kx$ for $K \neq 0$ represents parabolas. The question asks what the solution represents a family of, implying the general solution.

If we write $(y+3)^2 = Kx$, we have a family of parabolas. The horizontal line $y=-3$ corresponds to the vertex of the parabola at $x=0$. For $K=0$, we get $(y+3)^2=0$, which implies $y+3=0$, i.e., $y=-3$. So the case $K=0$ gives the horizontal line solution.

Thus, the general solution $(y+3)^2 = Kx$ represents a family of parabolas (including the horizontal line $y=-3$ when $K=0$).

The correct option is (C).

Given:

The differential equation $\frac{dy}{dx} (x \log x) + y = 2 \log x$.

To Find:

The integrating factor of the differential equation.

Solution:

The given differential equation is:

$\frac{dy}{dx} (x \log x) + y = 2 \log x$

This is a first-order linear differential equation. To find the integrating factor, we need to write it in the standard form $\frac{dy}{dx} + P(x)y = Q(x)$.

Divide the entire equation by $x \log x$ (assuming $x \log x \neq 0$, so $x \neq 1$ and $x > 0$):

$\frac{dy}{dx} + \frac{y}{x \log x} = \frac{2 \log x}{x \log x}$

$\frac{dy}{dx} + \frac{1}{x \log x} y = \frac{2}{x}$

Comparing this with the standard form, we identify $P(x)$ and $Q(x)$:

$P(x) = \frac{1}{x \log x}$

$Q(x) = \frac{2}{x}$

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

We need to evaluate the integral $\int P(x) dx = \int \frac{1}{x \log x} dx$.

Use the substitution method for the integral: Let $u = \log x$. Then $du = \frac{1}{x} dx$.

$\int \frac{1}{x \log x} dx = \int \frac{1}{\log x} \cdot \frac{1}{x} dx = \int \frac{1}{u} du$

$\int \frac{1}{u} du = \log |u| + C'$

Substitute back $u = \log x$:

$\int P(x) dx = \log |\log x| + C'$

The integrating factor is $IF = e^{\int P(x) dx}$. We can ignore the constant $C'$ when calculating the integrating factor.

$IF = e^{\log |\log x|}$

Using the property $e^{\log a} = a$, we get:

$IF = |\log x|$

Assuming $x>1$ so that $\log x > 0$, the integrating factor is $\log x$. If $0 < x < 1$, then $\log x < 0$, and the integrating factor is $-\log x$. However, usually, the absolute value is dropped, and the integrating factor is taken as $\log x$ (or $-\log x$, but the standard form $e^{\int P(x) dx}$ gives a specific sign if $P(x)$ is defined). Let's check the options.

The options suggest the integrating factor is $\log x$. Let's assume $x>1$ or proceed with $|\log x|$.

Let's check if $\log x$ works as the integrating factor (assuming $x>1$ where $\log x > 0$). Multiply the standard form equation by $\log x$:

$\log x \left(\frac{dy}{dx} + \frac{1}{x \log x} y\right) = \log x \left(\frac{2}{x}\right)$

$\log x \frac{dy}{dx} + \frac{1}{x} y = \frac{2 \log x}{x}$

The left side is $\frac{d}{dx}(y \cdot \log x)$ because $\frac{d}{dx}(y \log x) = \frac{dy}{dx} \log x + y \frac{d}{dx}(\log x) = \frac{dy}{dx} \log x + y \frac{1}{x}$, which matches.

So, the integrating factor is indeed $\log x$ (for $x>1$).

The correct option is (B).

Given:

The differential equation $\left( \frac{dy}{dx} \right)^2 - x \frac{dy}{dx} + y = 0$.

To Find:

Which of the given options is a solution to the differential equation.

Solution:

The given differential equation is a first-order non-linear equation:

$\left( \frac{dy}{dx} \right)^2 - x \frac{dy}{dx} + y = 0$

We need to test each option by substituting $y$ and $\frac{dy}{dx}$ into the differential equation.

Option (A): $y = 2$

If $y = 2$, then $\frac{dy}{dx} = \frac{d}{dx}(2) = 0$.

Substitute into the differential equation:

$(0)^2 - x(0) + 2 = 0$

$0 - 0 + 2 = 0$

$2 = 0$

This is false. So, $y=2$ is not a solution.

Option (B): $y = 2x$

If $y = 2x$, then $\frac{dy}{dx} = \frac{d}{dx}(2x) = 2$.

Substitute into the differential equation:

$(2)^2 - x(2) + 2x = 0$

$4 - 2x + 2x = 0$

$4 = 0$

This is false. So, $y=2x$ is not a solution.

Option (C): $y = 2x – 4$

If $y = 2x – 4$, then $\frac{dy}{dx} = \frac{d}{dx}(2x - 4) = 2 - 0 = 2$.

Substitute into the differential equation:

$(2)^2 - x(2) + (2x - 4) = 0$

$4 - 2x + 2x - 4 = 0$

$0 = 0$

This is true for all values of $x$. So, $y = 2x – 4$ is a solution.

Option (D): $y = 2x^2 – 4$

If $y = 2x^2 – 4$, then $\frac{dy}{dx} = \frac{d}{dx}(2x^2 - 4) = 4x - 0 = 4x$.

Substitute into the differential equation:

$(4x)^2 - x(4x) + (2x^2 - 4) = 0$

$16x^2 - 4x^2 + 2x^2 - 4 = 0$

$(16 - 4 + 2)x^2 - 4 = 0$

$14x^2 - 4 = 0$

This is not true for all values of $x$ (e.g., if $x=0$, $-4=0$ is false). So, $y = 2x^2 – 4$ is not a solution.

The only option that satisfies the differential equation is (C).

The correct option is (C).

Given:

Four functions of $x$ and $y$.

To Find:

Which of the given functions is not a homogeneous function of $x$ and $y$.

Solution:

A function $f(x, y)$ is said to be a homogeneous function of degree $n$ if $f(\lambda x, \lambda y) = \lambda^n f(x, y)$ for some real number $n$ and for all $\lambda \neq 0$ for which the function is defined.

We will check each option:

Option (A): $f(x, y) = x^2 + 2xy$

Evaluate $f(\lambda x, \lambda y)$:

$f(\lambda x, \lambda y) = (\lambda x)^2 + 2(\lambda x)(\lambda y)$

$f(\lambda x, \lambda y) = \lambda^2 x^2 + 2 \lambda^2 xy$

$f(\lambda x, \lambda y) = \lambda^2 (x^2 + 2xy)$

$f(\lambda x, \lambda y) = \lambda^2 f(x, y)$

This is a homogeneous function of degree 2.

Option (B): $f(x, y) = 2x – y$

Evaluate $f(\lambda x, \lambda y)$:

$f(\lambda x, \lambda y) = 2(\lambda x) – (\lambda y)$

$f(\lambda x, \lambda y) = 2 \lambda x – \lambda y$

$f(\lambda x, \lambda y) = \lambda (2x – y)$

$f(\lambda x, \lambda y) = \lambda^1 f(x, y)$

This is a homogeneous function of degree 1.

Option (C): $f(x, y) = \cos^2 \left( \frac{y}{x} \right) + \frac{y}{x}$

Evaluate $f(\lambda x, \lambda y)$:

$f(\lambda x, \lambda y) = \cos^2 \left( \frac{\lambda y}{\lambda x} \right) + \frac{\lambda y}{\lambda x}$

$f(\lambda x, \lambda y) = \cos^2 \left( \frac{y}{x} \right) + \frac{y}{x}$

$f(\lambda x, \lambda y) = 1 \cdot \left( \cos^2 \left( \frac{y}{x} \right) + \frac{y}{x} \right)$

$f(\lambda x, \lambda y) = \lambda^0 f(x, y)$

This is a homogeneous function of degree 0.

Option (D): $f(x, y) = \sin x – \cos y$

Evaluate $f(\lambda x, \lambda y)$:

$f(\lambda x, \lambda y) = \sin (\lambda x) – \cos (\lambda y)$

In general, $\sin (\lambda x) \neq \lambda^n \sin x$ and $\cos (\lambda y) \neq \lambda^n \cos y$ for any fixed power $n$. Thus, $f(\lambda x, \lambda y) = \sin (\lambda x) – \cos (\lambda y)$ cannot be expressed in the form $\lambda^n (\sin x – \cos y)$.

For example, if $\lambda = 2$, $f(2x, 2y) = \sin(2x) - \cos(2y)$. Is this equal to $2^n (\sin x - \cos y)$ for some $n$? No.

Therefore, this function is not homogeneous.

The correct option is (D).

Given:

The differential equation $\frac{dx}{x} + \frac{dy}{y} = 0$.

To Find:

The solution of the differential equation.

Solution:

The given differential equation is already in a separable form:

$\frac{dx}{x} + \frac{dy}{y} = 0$

Integrate both sides of the equation:

$\int \frac{1}{x} dx + \int \frac{1}{y} dy = \int 0$

$\log |x| + \log |y| = C'$

where $C'$ is the constant of integration.

Using the logarithm property $\log a + \log b = \log (ab)$, we can combine the terms on the left side:

$\log |xy| = C'$

To remove the logarithm, exponentiate both sides using the base $e$:

$|xy| = e^{C'}$

Let $e^{C'} = c$, where $c$ is a positive arbitrary constant.

$|xy| = c$

This means $xy = \pm c$. Let $C = \pm c$. Since $c$ is a positive arbitrary constant, $C$ is an arbitrary non-zero constant.

The general solution is:

$xy = C$

We should also consider cases where $x=0$ or $y=0$. If $x=0$, the term $\frac{dx}{x}$ is undefined. If $y=0$, the term $\frac{dy}{y}$ is undefined. The original equation is $y dx + x dy = 0$ (after multiplying by $xy$), which is the differential of $xy=C$. The solution $xy=C$ covers the cases where $x \neq 0$ and $y \neq 0$. The solution $xy=C$ also includes the lines $x=0$ (when $C=0$) and $y=0$ (when $C=0$).

Looking at the options, the constant is denoted by $c$. So the solution is $xy = c$, where $c$ is an arbitrary constant.

The correct option is (C).

Given:

The differential equation $x \frac{dy}{dx} + 2y = x^2$.

To Find:

The solution of the differential equation.

Solution:

The given differential equation is:

$x \frac{dy}{dx} + 2y = x^2$

This is a first-order linear differential equation. We can write it in the standard form $\frac{dy}{dx} + P(x)y = Q(x)$ by dividing by $x$ (assuming $x \neq 0$):

$\frac{dy}{dx} + \frac{2}{x}y = x$

Here, $P(x) = \frac{2}{x}$ and $Q(x) = x$.

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

Calculate the integral of $P(x)$:

$\int P(x) dx = \int \frac{2}{x} dx = 2 \int \frac{1}{x} dx = 2 \log |x| = \log |x|^2 = \log (x^2)$

The integrating factor is:

$IF = e^{\int P(x) dx} = e^{\log (x^2)} = x^2$

Multiply the standard form of the differential equation by the integrating factor $x^2$:

$x^2 \left(\frac{dy}{dx} + \frac{2}{x}y\right) = x^2 (x)$

$x^2 \frac{dy}{dx} + 2xy = x^3$

The left side of the equation is the derivative of the product of $y$ and the integrating factor, i.e., $\frac{d}{dx}(y \cdot x^2)$.

So the equation becomes:

$\frac{d}{dx}(yx^2) = x^3$

Integrate both sides with respect to $x$:

$\int \frac{d}{dx}(yx^2) dx = \int x^3 dx$

$yx^2 = \frac{x^{3+1}}{3+1} + C$

$yx^2 = \frac{x^4}{4} + C$

where $C$ is the constant of integration.

Solve for $y$ by dividing both sides by $x^2$ (for $x \neq 0$):

$y = \frac{\frac{x^4}{4} + C}{x^2}$

$y = \frac{x^4}{4x^2} + \frac{C}{x^2}$

$y = \frac{x^2}{4} + \frac{C}{x^2}$

Compare this solution with the given options. We can rewrite option (D):

(D) $y = \frac{x^4 + c}{4x^2} = \frac{x^4}{4x^2} + \frac{c}{4x^2} = \frac{x^2}{4} + \frac{c}{4x^2}$

This matches our derived solution $y = \frac{x^2}{4} + \frac{C}{x^2}$ if the arbitrary constant $C$ in our solution is equal to $\frac{c}{4}$ from option (D). Since $c$ in option (D) is an arbitrary constant, $\frac{c}{4}$ is also an arbitrary constant. Thus, option (D) represents the general solution.

The correct option is (D).

Answer for (i): The differential equation representing the family of parabolas $y^2 = 4ax$ involves one arbitrary constant ($a$). Differentiating once with respect to $x$, we get $2y \frac{dy}{dx} = 4a$. Substituting $4a$ from the original equation, we get $2y \frac{dy}{dx} = \frac{y^2}{x}$, which simplifies to $2x \frac{dy}{dx} = y$. This is a first-order differential equation. Therefore, the order is 1.

Answer for (ii): The given differential equation is $\left( \frac{dy}{dx} \right)^3 + \left( \frac{d^2y}{dx^2} \right)^2 = 0$. The highest order derivative present is $\frac{d^2y}{dx^2}$, which has an order of 2. The power of this highest order derivative is 2. Therefore, the degree of the differential equation is 2.

Answer for (iii): A particular solution of a differential equation is obtained by assigning specific values to the arbitrary constants in the general solution. Consequently, a particular solution contains 0 arbitrary constants.

Answer for (iv): A function $F(x,y)$ is homogeneous of degree $n$ if $F(tx, ty) = t^n F(x,y)$. For the given function $F(x,y) = \frac{\sqrt{x^2+y^2} + y}{x}$, we have $F(tx, ty) = \frac{\sqrt{(tx)^2+(ty)^2} + ty}{tx} = \frac{\sqrt{t^2(x^2+y^2)} + ty}{tx} = \frac{t\sqrt{x^2+y^2} + ty}{tx} = \frac{t(\sqrt{x^2+y^2} + y)}{tx} = \frac{\sqrt{x^2+y^2} + y}{x} = t^0 F(x,y)$. Therefore, the function is homogeneous of degree 0.

Answer for (v): The given differential equation is $\frac{dx}{dy} = \frac{x^2 \log \left( \frac{x}{y} \right) - x^2}{xy \log \left( \frac{x}{y} \right)}$. The right-hand side can be written as $\frac{x^2 (\log(x/y) - 1)}{xy \log(x/y)} = \frac{x (\log(x/y) - 1)}{y \log(x/y)} = \frac{(x/y) (\log(x/y) - 1)}{\log(x/y)}$. Since the right-hand side is a function of $\frac{x}{y}$, an appropriate substitution is to let $v = x/y$, or $x = vy$, where $v$ is a function of $y$. The substitution is $x/y = v$.

Answer for (vi): The given differential equation is $x \frac{dy}{dx} − y = \sin x$. To write it in the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$, we divide by $x$: $\frac{dy}{dx} - \frac{1}{x}y = \frac{\sin x}{x}$. Here, $P(x) = - \frac{1}{x}$. The integrating factor (IF) is given by $e^{\int P(x) dx}$. IF $= e^{\int -\frac{1}{x} dx} = e^{-\log|x|} = e^{\log|x|^{-1}} = e^{\log|\frac{1}{x}|} = |\frac{1}{x}|$. Assuming $x > 0$, the integrating factor is $1/x$.

Answer for (vii): The given differential equation is $\frac{dy}{dx} = e^{x-y}$. We can rewrite this as $\frac{dy}{dx} = e^x e^{-y} = \frac{e^x}{e^y}$. This is a separable variable equation. Separating the variables, we get $e^y dy = e^x dx$. Integrating both sides, $\int e^y dy = \int e^x dx$. This gives $e^y = e^x + C$, where $C$ is the arbitrary constant. The general solution is $e^y = e^x + C$.

Answer for (viii): The given differential equation is $\frac{dy}{dx} + \frac{y}{x} = 1$. This is a linear first-order differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = \frac{1}{x}$ and $Q(x) = 1$. The integrating factor (IF) is $e^{\int P(x) dx} = e^{\int \frac{1}{x} dx} = e^{\log|x|} = |x|$. Assuming $x > 0$, IF $= x$. The general solution is given by $y \cdot \text{IF} = \int Q(x) \cdot \text{IF} \; dx + C$. $y \cdot x = \int 1 \cdot x \; dx + C$. $xy = \frac{x^2}{2} + C$. The general solution is $y = \frac{x}{2} + \frac{C}{x}$.

Answer for (ix): The given family of curves is $y = A \sin x + B \cos x$. It contains two arbitrary constants, $A$ and $B$. We need to differentiate twice to eliminate them. $\frac{dy}{dx} = A \cos x - B \sin x$. $\frac{d^2y}{dx^2} = -A \sin x - B \cos x = -(A \sin x + B \cos x) = -y$. Rearranging, we get $\frac{d^2y}{dx^2} + y = 0$. The differential equation is $\frac{d^2y}{dx^2} + y = 0$.

Answer for (x): The given equation is $\left( \frac{e^{−2\sqrt{x}}}{\sqrt{x}} − \frac{y}{\sqrt{x}} \right) \frac{dy}{dx} = 1$. We need to write this in the form $\frac{dy}{dx} + Py = Q$. Divide the equation by the term multiplying $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{1}{\frac{e^{−2\sqrt{x}}}{\sqrt{x}} − \frac{y}{\sqrt{x}}} = \frac{\sqrt{x}}{e^{−2\sqrt{x}} - y}$. Rearranging to isolate terms: $\frac{dy}{dx} (e^{-2\sqrt{x}} - y) = \sqrt{x}$. $e^{-2\sqrt{x}} \frac{dy}{dx} - y \frac{dy}{dx} = \sqrt{x}$. This equation as given cannot be directly put in the form $\frac{dy}{dx} + Py = Q$ where $P$ and $Q$ are functions of $x$ alone, due to the presence of the $y \frac{dy}{dx}$ term. However, based on standard problems of this type and likely intended meaning (as found in common sources), the intended equation might be one that rearranges to a standard linear form. Assuming the equation is intended to be linear in $y$, and based on typical solutions for such problems, the coefficient $P$ in the form $\frac{dy}{dx} + Py = Q$ is $1/\sqrt{x}$.

(i) The family of ellipses having centre at origin and foci on the x-axis is given by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. This equation has two arbitrary constants, $a^2$ and $b^2$. To form the differential equation, we need to differentiate twice to eliminate these two constants. Therefore, the order of the differential equation is 2. Statement is True.

(ii) The given differential equation is $\sqrt{1 + \frac{d^2y}{dx^2}} = x + \frac{dy}{dx}$. To find the degree, we must make it a polynomial in derivatives. Squaring both sides, we get $1 + \frac{d^2y}{dx^2} = \left( x + \frac{dy}{dx} \right)^2 = x^2 + 2x \frac{dy}{dx} + \left( \frac{dy}{dx} \right)^2$. The highest order derivative is $\frac{d^2y}{dx^2}$, and its power is 1. Since it can be written as a polynomial in the derivatives, the degree is defined and is 1. Statement is False.

(iii) The differential equation is $\frac{dy}{dx} + y = 5$. This is in the form $\frac{dy}{dx} + P(x)y = Q(x)$ with $P(x) = 1$ and $Q(x) = 5$. Also, it can be written as $\frac{dy}{dx} = 5 - y$, which is a variable separable equation: $\frac{dy}{5-y} = dx$. So, it can be solved using the variable separable method. Statement is True.

(iv) Let $F(x, y) = \frac{y \cos \left( \frac{y}{x} \right) + x}{x \cos \left( \frac{y}{x} \right)}$. Consider $F(tx, ty) = \frac{ty \cos \left( \frac{ty}{tx} \right) + tx}{tx \cos \left( \frac{ty}{tx} \right)} = \frac{t \left( y \cos \left( \frac{y}{x} \right) + x \right)}{t \left( x \cos \left( \frac{y}{x} \right) \right)} = \frac{y \cos \left( \frac{y}{x} \right) + x}{x \cos \left( \frac{y}{x} \right)} = t^0 F(x, y)$. Since $F(tx, ty) = t^0 F(x, y)$, the function is homogeneous of degree 0. The statement says it is not a homogeneous function. Statement is False.

(v) Let $F(x, y) = \frac{x^2 + y^2}{x − y}$. Consider $F(tx, ty) = \frac{(tx)^2 + (ty)^2}{tx − ty} = \frac{t^2 x^2 + t^2 y^2}{t(x − y)} = \frac{t^2(x^2 + y^2)}{t(x − y)} = t \frac{x^2 + y^2}{x − y} = t^1 F(x, y)$. Since $F(tx, ty) = t^1 F(x, y)$, the function is homogeneous of degree 1. Statement is True.

(vi) The given differential equation is $\frac{dy}{dx} − y = \cos x$. This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = -1$. The integrating factor (IF) is $e^{\int P(x) dx} = e^{\int -1 dx} = e^{-x}$. The statement says the integrating factor is $e^x$. Statement is False.

(vii) The given differential equation is $x(1 + y^2)dx + y(1 + x^2)dy = 0$. This is a variable separable equation. Dividing by $(1+x^2)(1+y^2)$, we get $\frac{x}{1+x^2} dx + \frac{y}{1+y^2} dy = 0$. Integrating both sides, $\int \frac{x}{1+x^2} dx + \int \frac{y}{1+y^2} dy = C$. This gives $\frac{1}{2} \log(1+x^2) + \frac{1}{2} \log(1+y^2) = C$, which simplifies to $\log((1+x^2)(1+y^2)) = 2C$. Let $2C = \log k$. Then $\log((1+x^2)(1+y^2)) = \log k$, implying $(1+x^2)(1+y^2) = k$. Statement is True.

(viii) The given differential equation is $\frac{dy}{dx} + y \sec x = \tan x$. This is a linear differential equation with $P(x) = \sec x$. The integrating factor is $e^{\int \sec x dx} = e^{\log|\sec x + \tan x|} = |\sec x + \tan x|$. Assuming $\sec x + \tan x > 0$, the integrating factor is $\sec x + \tan x$. The general solution is $y(\sec x + \tan x) = \int \tan x (\sec x + \tan x) dx + C = \int (\sec x \tan x + \tan^2 x) dx + C = \int \sec x \tan x dx + \int (\sec^2 x - 1) dx + C = \sec x + \tan x - x + C$. The provided solution is $y (\sec x – \tan x) = \sec x – \tan x + x + k$, which does not match the calculated general solution. Statement is False.

(ix) The given relation is $x + y = \tan^{-1} y$. We differentiate with respect to $x$ to find $\frac{dy}{dx}$: $1 + \frac{dy}{dx} = \frac{1}{1+y^2} \frac{dy}{dx}$. Rearranging, $1 = \frac{dy}{dx} \left( \frac{1}{1+y^2} - 1 \right) = \frac{dy}{dx} \left( \frac{1 - (1+y^2)}{1+y^2} \right) = \frac{dy}{dx} \left( \frac{-y^2}{1+y^2} \right)$. So, $\frac{dy}{dx} = - \frac{1+y^2}{y^2}$. Substitute this into the differential equation $y^2 \frac{dy}{dx} + y^2 + 1 = 0$: $y^2 \left( - \frac{1+y^2}{y^2} \right) + y^2 + 1 = -(1+y^2) + y^2 + 1 = -1 - y^2 + y^2 + 1 = 0$. Since substituting the derivatives and $y$ from the given relation satisfies the differential equation, the relation is a solution. Statement is True.

(x) If $y = x$ is a solution to the differential equation $\frac{d^2y}{dx^2} − x^2 \frac{dy}{dx} + xy = x$, then substituting $y=x$, $\frac{dy}{dx} = 1$, and $\frac{d^2y}{dx^2} = 0$ into the equation should result in an identity. Substituting, we get $0 - x^2(1) + x(x) = x$, which simplifies to $-x^2 + x^2 = x$, or $0 = x$. This is only true for $x=0$, not for all values of $x$ in the domain. Therefore, $y=x$ is not a solution. Statement is False.

Given the differential equation:

$\frac{dy}{dx} = 2^{y-x}$

We can rewrite the right-hand side using the properties of exponents:

$\frac{dy}{dx} = 2^y \cdot 2^{-x}$

$\frac{dy}{dx} = \frac{2^y}{2^x}$

This is a variable separable differential equation. We can separate the terms involving $y$ and $dy$ from the terms involving $x$ and $dx$:

$\frac{dy}{2^y} = \frac{dx}{2^x}$

$2^{-y} dy = 2^{-x} dx$

Integrate both sides of the separated equation:

$\int 2^{-y} dy = \int 2^{-x} dx$

Recall the integral of $a^u du$ is $\frac{a^u}{\ln a} + C$. Applying this formula to both sides:

$\int 2^{-y} dy = \frac{2^{-y}}{\ln 2} \cdot (-1) + C_1 = -\frac{2^{-y}}{\ln 2} + C_1$

$\int 2^{-x} dx = \frac{2^{-x}}{\ln 2} \cdot (-1) + C_2 = -\frac{2^{-x}}{\ln 2} + C_2$

Equating the results of the integrals:

$-\frac{2^{-y}}{\ln 2} = -\frac{2^{-x}}{\ln 2} + C$

Multiply the entire equation by $-\ln 2$ to simplify:

$2^{-y} = 2^{-x} - C \ln 2$

So, the general solution is:

$2^{-y} = 2^{-x} + K$

or

$\frac{1}{2^y} = \frac{1}{2^x} + K$

Final Answer:

The solution of the given differential equation is $\mathbf{2^{-y} = 2^{-x} + K}$, where $\mathbf{K}$ is an arbitrary constant.

Given: Family of all non-vertical lines in a plane.

To Find: The differential equation representing this family.

Solution:

The general equation of a non-vertical line in a plane is given by:

$y = mx + c$

This equation contains two arbitrary constants ($m$ and $c$). To find the differential equation, we need to eliminate these two constants by differentiation. The order of the differential equation will be equal to the number of arbitrary constants, which is 2.

Differentiate the equation ($y = mx + c$) with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(mx + c)$

$\frac{dy}{dx} = m \frac{d}{dx}(x) + \frac{d}{dx}(c)$

$\frac{dy}{dx} = m(1) + 0$

Now, differentiate the equation (i) with respect to $x$ to eliminate the constant $m$:

$\frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(m)$

$\frac{d^2y}{dx^2} = 0$

Equation (ii) is a differential equation involving derivatives of $y$ with respect to $x$. It does not contain any arbitrary constants from the original equation $y = mx + c$. The highest order derivative is $\frac{d^2y}{dx^2}$, so the order of this differential equation is 2.

Final Answer:

The differential equation of all non-vertical lines in a plane is $\frac{d^2y}{dx^2} = 0$.

Given:

The differential equation $\frac{dy}{dx} = e^{−2y}$.

The initial condition $y = 0$ when $x = 5$.

To Find:

The value of $x$ when $y = 3$.

Solution:

The given differential equation is:

$\frac{dy}{dx} = e^{−2y}$

This is a variable separable differential equation. We can separate the variables $x$ and $y$:

$dy = e^{-2y} dx$

$\frac{dy}{e^{-2y}} = dx$

$e^{2y} dy = dx$

Integrate both sides of the equation:

$\int e^{2y} dy = \int dx$

Performing the integration:

$\frac{e^{2y}}{2} = x + C$

Now, we use the given initial condition $y = 0$ when $x = 5$ to find the value of the constant $C$. Substitute $x=5$ and $y=0$ into the general solution:

$\frac{e^{2(0)}}{2} = 5 + C$

$\frac{e^0}{2} = 5 + C$

$\frac{1}{2} = 5 + C$

Solve for $C$:

$C = \frac{1}{2} - 5$

$C = \frac{1 - 10}{2}$

$C = -\frac{9}{2}$

Substitute the value of $C$ back into the general solution to obtain the particular solution:

$\frac{e^{2y}}{2} = x - \frac{9}{2}$

Finally, we need to find the value of $x$ when $y = 3$. Substitute $y=3$ into the particular solution:

$\frac{e^{2(3)}}{2} = x - \frac{9}{2}$

$\frac{e^6}{2} = x - \frac{9}{2}$

Solve for $x$:

$x = \frac{e^6}{2} + \frac{9}{2}$

$x = \frac{e^6 + 9}{2}$

Final Answer:

The value of $x$ when $y=3$ is $\frac{e^6 + 9}{2}$.

Given:

The differential equation $(x^2 – 1) \frac{dy}{dx} + 2xy = \frac{1}{x^2 − 1}$.

To Solve:

Find the general solution of the given differential equation.

Solution:

The given differential equation is $(x^2 – 1) \frac{dy}{dx} + 2xy = \frac{1}{x^2 − 1}$.

This equation is a linear first-order differential equation. To write it in the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we divide the entire equation by $(x^2 - 1)$, assuming $x^2 - 1 \neq 0$.

$\frac{dy}{dx} + \frac{2xy}{x^2 − 1} = \frac{\frac{1}{x^2 − 1}}{x^2 − 1}$

$\frac{dy}{dx} + \frac{2x}{x^2 − 1}y = \frac{1}{(x^2 − 1)^2}$

Comparing equation (i) with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we identify $P(x)$ and $Q(x)$:

$P(x) = \frac{2x}{x^2 − 1}$

$Q(x) = \frac{1}{(x^2 − 1)^2}$

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

First, we calculate the integral $\int P(x) dx$:

$\int P(x) dx = \int \frac{2x}{x^2 − 1} dx$

Let $u = x^2 - 1$. Then $du = 2x dx$. The integral becomes:

$\int \frac{du}{u} = \ln|u| = \ln|x^2 − 1|$

Now, calculate the integrating factor:

IF $= e^{\int P(x) dx} = e^{\ln|x^2 − 1|}$

IF $= |x^2 − 1|$

For simplicity in the general solution, we can use IF $= x^2 - 1$ (the absolute value will be absorbed by the arbitrary constant later, assuming $x^2-1$ keeps a consistent sign over the interval of interest where $x^2-1 \neq 0$).

The general solution of a linear first-order differential equation is given by:

$y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) \; dx + C$

Substitute $Q(x)$ and IF into equation (iii):

$y(x^2 − 1) = \int \frac{1}{(x^2 − 1)^2} \cdot (x^2 − 1) \; dx + C$

$y(x^2 − 1) = \int \frac{1}{x^2 − 1} \; dx + C$

The integral $\int \frac{1}{x^2 − 1} dx$ is a standard integral:

$\int \frac{1}{x^2 − a^2} dx = \frac{1}{2a} \ln\left|\frac{x-a}{x+a}\right| + C'$

In our case, $a^2 = 1$, so $a = 1$.

$\int \frac{1}{x^2 − 1} dx = \frac{1}{2(1)} \ln\left|\frac{x-1}{x+1}\right| = \frac{1}{2} \ln\left|\frac{x-1}{x+1}\right|$

Substitute the result of the integral back into the general solution equation:

$y(x^2 − 1) = \frac{1}{2} \ln\left|\frac{x-1}{x+1}\right| + C$

We can express $y$ explicitly:

$y = \frac{1}{x^2 − 1} \left( \frac{1}{2} \ln\left|\frac{x-1}{x+1}\right| + C \right)$

Final Answer:

The general solution of the differential equation is $y(x^2 − 1) = \frac{1}{2} \ln\left|\frac{x-1}{x+1}\right| + C$ or $y = \frac{1}{x^2 − 1} \left( \frac{1}{2} \ln\left|\frac{x-1}{x+1}\right| + C \right)$, where $\mathbf{C}$ is the arbitrary constant.

Given:

The differential equation $\frac{dy}{dx} + 2xy = y$.

To Solve:

Find the general solution of the given differential equation.

Solution:

The given differential equation is:

$\frac{dy}{dx} + 2xy = y$

We can rearrange the equation to group the terms involving $y$:

$\frac{dy}{dx} = y - 2xy$

$\frac{dy}{dx} = y(1 - 2x)$

This is a variable separable differential equation. We can separate the terms involving $y$ and $dy$ from the terms involving $x$ and $dx$, assuming $y \neq 0$:

$\frac{dy}{y} = (1 - 2x) dx$

Integrate both sides of the separated equation:

$\int \frac{dy}{y} = \int (1 - 2x) dx$

Perform the integration:

$\int \frac{dy}{y} = \ln|y|$

$\int (1 - 2x) dx = \int 1 \, dx - \int 2x \, dx = x - 2 \frac{x^2}{2} + C = x - x^2 + C$

Equating the results of the integrals:

$\ln|y| = x - x^2 + C$

To solve for $y$, exponentiate both sides with base $e$:

$|y| = e^{x - x^2 + C}$

$|y| = e^{x - x^2} \cdot e^C$

Let $A = \pm e^C$. Since $C$ is an arbitrary constant, $e^C$ is a positive constant, and $A$ can be any non-zero constant. If we also consider the case $y=0$ (which is a solution to the original differential equation $\frac{dy}{dx} = y(1-2x)$ as $\frac{d(0)}{dx} = 0$ and $0(1-2x) = 0$), we can include $A=0$. Therefore, $A$ can be any real constant.

$y = A e^{x - x^2}$

Final Answer:

The general solution of the differential equation is $y = A e^{x - x^2}$, where $\mathbf{A}$ is an arbitrary constant.

Given:

The differential equation $\frac{dy}{dx} + ay = e^{mx}$.

To Find:

The general solution of the given differential equation.

Solution:

The given differential equation is $\frac{dy}{dx} + ay = e^{mx}$.

This is a linear first-order differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = a$ and $Q(x) = e^{mx}$.

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

Calculate the integral $\int P(x) dx$:

$\int P(x) dx = \int a \, dx = ax$

Calculate the integrating factor:

IF $= e^{\int P(x) dx} = e^{ax}$

The general solution of a linear first-order differential equation is given by:

$y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) \; dx + C$

Substitute $Q(x)$ and IF into the general solution formula:

$y \cdot e^{ax} = \int e^{mx} \cdot e^{ax} \; dx + C$

$y e^{ax} = \int e^{(m+a)x} \; dx + C$

We need to evaluate the integral $\int e^{(m+a)x} \; dx$. There are two cases depending on the value of $m+a$.

Case 1: $m + a \neq 0$

The integral is:

$\int e^{(m+a)x} \; dx = \frac{e^{(m+a)x}}{m+a}$

Substitute this back into the general solution equation:

$y e^{ax} = \frac{e^{(m+a)x}}{m+a} + C$

Divide by $e^{ax}$ to solve for $y$:

$y = \frac{e^{(m+a)x}}{(m+a)e^{ax}} + \frac{C}{e^{ax}}$

$y = \frac{e^{mx} e^{ax}}{(m+a)e^{ax}} + C e^{-ax}$

$\mathbf{y = \frac{e^{mx}}{m+a} + C e^{-ax}}$

Case 2: $m + a = 0$ (i.e., $m = -a$)

The integral becomes:

$\int e^{(m+a)x} \; dx = \int e^{0x} \; dx = \int 1 \; dx = x$

Substitute this back into the general solution equation:

$y e^{ax} = x + C$

Divide by $e^{ax}$ to solve for $y$:

$y = \frac{x}{e^{ax}} + \frac{C}{e^{ax}}$

$\mathbf{y = x e^{-ax} + C e^{-ax}}$

Final Answer:

The general solution of the differential equation is:

If $m+a \neq 0$, then $y = \frac{e^{mx}}{m+a} + C e^{-ax}$.

If $m+a = 0$, then $y = x e^{-ax} + C e^{-ax}$.

Here, $\mathbf{C}$ is the arbitrary constant of integration.

Given:

The differential equation $\frac{dy}{dx} + 1 = e^{x+y}$.

To Solve:

Find the general solution of the given differential equation.

Solution:

The given differential equation is:

$\frac{dy}{dx} + 1 = e^{x+y}$

We can rewrite the equation as:

$\frac{dy}{dx} = e^{x+y} - 1$

Let's use the substitution $z = x+y$.

Differentiating $z$ with respect to $x$, we get:

$\frac{dz}{dx} = \frac{d}{dx}(x+y)$

$\frac{dz}{dx} = 1 + \frac{dy}{dx}$

From the given differential equation, $\frac{dy}{dx} = e^{x+y} - 1$. Substitute this into the differentiated substitution equation:

$\frac{dz}{dx} = 1 + (e^{x+y} - 1)$

$\frac{dz}{dx} = e^{x+y}$

Substitute $z = x+y$ back into this equation:

$\frac{dz}{dx} = e^z$

This is a variable separable differential equation. Separate the variables $z$ and $x$:

$\frac{dz}{e^z} = dx$

$e^{-z} dz = dx$

Integrate both sides of the separated equation:

$\int e^{-z} dz = \int dx$

Perform the integration:

$-e^{-z} = x + C$

Substitute back $z = x+y$ into the integrated equation:

$-e^{-(x+y)} = x + C$

We can rearrange this result in different forms. One possible form is:

$e^{-(x+y)} + x + C = 0$

Final Answer:

The general solution of the differential equation is $-e^{-(x+y)} = x + C$ or $e^{-(x+y)} + x + C = 0$, where $\mathbf{C}$ is the arbitrary constant.

Given:

The differential equation $y\;dx – x\;dy = x^2y\;dx$.

To Solve:

Find the general solution of the given differential equation.

Solution:

The given differential equation is $y\;dx – x\;dy = x^2y\;dx$.

We can rearrange the terms to group the $dx$ terms:

$y\;dx – x^2y\;dx = x\;dy$

$(y - x^2y)\;dx = x\;dy$

$y(1 - x^2)\;dx = x\;dy$

This is a variable separable differential equation. We can separate the variables $x$ and $y$ by dividing both sides by $xy(1-x^2)$, assuming $x \neq 0$, $y \neq 0$, and $1-x^2 \neq 0$ (i.e., $x \neq \pm 1$). Dividing by $xy$: (Note: the case $y=0$ will be checked later)

$\frac{y(1 - x^2)}{xy} dx = \frac{x dy}{xy}$

$\frac{1 - x^2}{x} dx = \frac{dy}{y}$

$\left(\frac{1}{x} - x\right) dx = \frac{dy}{y}$

Integrate both sides of the separated equation:

$\int \left(\frac{1}{x} - x\right) dx = \int \frac{dy}{y}$

Perform the integration:

$\int \frac{1}{x} dx - \int x dx = \int \frac{dy}{y}$

$\ln|x| - \frac{x^2}{2} = \ln|y| + C'$

Rearrange the equation to isolate the terms involving $\ln$:

$\ln|x| - \ln|y| = C' + \frac{x^2}{2}$

$\ln\left|\frac{x}{y}\right| = C' + \frac{x^2}{2}$

Exponentiate both sides with base $e$:

$\left|\frac{x}{y}\right| = e^{C' + \frac{x^2}{2}}$

$\left|\frac{x}{y}\right| = e^{C'} \cdot e^{\frac{x^2}{2}}$

Remove the absolute value by introducing a new constant $A = \pm e^{C'}$. Since $e^{C'} > 0$, $A$ is a non-zero constant.

$\frac{x}{y} = A e^{\frac{x^2}{2}}$

Solve for $y$:

$y = \frac{x}{A e^{\frac{x^2}{2}}}$

$y = \frac{1}{A} x e^{-\frac{x^2}{2}}$

Let $K = \frac{1}{A}$. Since $A$ is a non-zero constant, $K$ is also a non-zero constant.

$y = K x e^{-\frac{x^2}{2}}$

Now, let's consider the case $y=0$. Substituting $y=0$ into the original differential equation $y\;dx – x\;dy = x^2y\;dx$, we get:

$0 \cdot dx - x \cdot d(0) = x^2 \cdot 0 \cdot dx$

$0 - 0 = 0$

$0 = 0$

Let's check if the solution $y=0$ is included in the general solution $y = K x e^{-\frac{x^2}{2}}$. If we set $K=0$ in the general solution, we get $y = 0 \cdot x \cdot e^{-\frac{x^2}{2}} = 0$. Therefore, the solution $y=0$ is included in the general solution if we allow $K$ to be any real constant (including zero).

Final Answer:

The general solution of the differential equation is $y = K x e^{-\frac{x^2}{2}}$, where $\mathbf{K}$ is an arbitrary constant.

Given:

The differential equation $\frac{dy}{dx} = 1 + x + y^2 + xy^2$.

The initial condition $y = 0$ when $x = 0$.

To Solve:

Find the particular solution of the given differential equation that satisfies the initial condition.

Solution:

The given differential equation is:

$\frac{dy}{dx} = 1 + x + y^2 + xy^2$

We can factor the right-hand side of the equation:

$\frac{dy}{dx} = (1 + x) + y^2(1 + x)$

$\frac{dy}{dx} = (1 + x)(1 + y^2)$

This is a variable separable differential equation. We can separate the terms involving $y$ and $dy$ from the terms involving $x$ and $dx$:

$\frac{dy}{1 + y^2} = (1 + x) dx$

Integrate both sides of the separated equation:

$\int \frac{dy}{1 + y^2} = \int (1 + x) dx$

Perform the integration:

$\tan^{-1} y = x + \frac{x^2}{2} + C$

Now, we use the given initial condition $y = 0$ when $x = 0$ to find the value of the constant $C$. Substitute $x=0$ and $y=0$ into the general solution:

$\tan^{-1}(0) = 0 + \frac{0^2}{2} + C$

$0 = 0 + 0 + C$

$C = 0$

Substitute the value of $C=0$ back into the general solution to obtain the particular solution:

$\tan^{-1} y = x + \frac{x^2}{2}$

Final Answer:

The particular solution of the differential equation that satisfies the given initial condition is $\tan^{-1} y = x + \frac{x^2}{2}$.

Given:

The differential equation $(x + 2y^3) \frac{dy}{dx} = y$.

To Solve:

Find the general solution of the given differential equation.

Solution:

The given differential equation is:

$(x + 2y^3) \frac{dy}{dx} = y$

We can rewrite the differential equation by taking the reciprocal of $\frac{dy}{dx}$ to get $\frac{dx}{dy}$, assuming $y \neq 0$ and $x + 2y^3 \neq 0$:

$\frac{dx}{dy} = \frac{x + 2y^3}{y}$

Separate the terms on the right-hand side:

$\frac{dx}{dy} = \frac{x}{y} + \frac{2y^3}{y}$

$\frac{dx}{dy} = \frac{x}{y} + 2y^2$

Rearrange the equation to write it in the standard form of a linear first-order differential equation in $x$ with respect to $y$: $\frac{dx}{dy} + P(y)x = Q(y)$

$\frac{dx}{dy} - \frac{1}{y} x = 2y^2$

The integrating factor (IF) is given by $e^{\int P(y) dy}$.

Calculate the integral $\int P(y) dy$:

$\int P(y) dy = \int -\frac{1}{y} dy = -\ln|y|$

Calculate the integrating factor:

IF $= e^{\int P(y) dy} = e^{-\ln|y|} = e^{\ln|y|^{-1}} = e^{\ln|\frac{1}{y}|} = \left|\frac{1}{y}\right|$

For simplicity in the general solution, we can use IF $= \frac{1}{y}$ (the absolute value will be absorbed by the arbitrary constant later, assuming $y \neq 0$).

The general solution of a linear first-order differential equation in $x$ is given by:

$x \cdot (\text{IF}) = \int Q(y) \cdot (\text{IF}) \; dy + C$

Substitute $Q(y)$ and IF into the general solution formula:

$x \cdot \frac{1}{y} = \int (2y^2) \cdot \frac{1}{y} \; dy + C$

$\frac{x}{y} = \int 2y \; dy + C$

Evaluate the integral $\int 2y \; dy$:

$\int 2y \; dy = 2 \frac{y^{1+1}}{1+1} = 2 \frac{y^2}{2} = y^2$

Substitute the result of the integral back into the general solution equation:

$\frac{x}{y} = y^2 + C$

Solve for $x$ by multiplying both sides by $y$:

$x = y(y^2 + C)$

$x = y^3 + Cy$

We should also check if $y=0$ is a solution to the original equation. Substituting $y=0$ and $\frac{dy}{dx}=0$ into $(x + 2y^3) \frac{dy}{dx} = y$ gives $(x + 0) \cdot 0 = 0$, which is $0=0$. So $y=0$ is a solution. However, the general solution $x = y^3 + Cy$ represents curves in the plane. For $y=0$, we get $x=0$, meaning the origin $(0,0)$ is on every curve in the family. The line $y=0$ is not a part of this family unless $x=0$ everywhere, which is not the case for $y=0$. Thus, $y=0$ is a singular solution not included in the general solution obtained by the method for $y \neq 0$. However, standard practice often provides the general solution obtained via the standard method.

Final Answer:

The general solution of the differential equation is $x = y^3 + Cy$, where $\mathbf{C}$ is an arbitrary constant.

Given:

The differential equation $\left( \frac{2 + \sin x}{1 + y} \right) \frac{dy}{dx} = - \cos x$.

The initial condition $y(0) = 1$.

To Find:

The value of $y \left( \frac{\pi}{2} \right)$.

Solution:

The given differential equation is:

This is a variable separable differential equation. We can separate the variables $x$ and $y$ by multiplying both sides by $\frac{1+y}{2 + \sin x}$, assuming $1+y \neq 0$ and $2 + \sin x \neq 0$:

$\frac{dy}{1+y} = - \cos x \cdot \frac{dx}{2 + \sin x}$

$\frac{dy}{1+y} = \frac{- \cos x}{2 + \sin x} dx$

Integrate both sides of the separated equation:

$\int \frac{dy}{1+y} = \int \frac{- \cos x}{2 + \sin x} dx$

For the left-hand side integral, $\int \frac{dy}{1+y} = \ln|1+y|$.

For the right-hand side integral, let $u = 2 + \sin x$. Then $du = \cos x \; dx$. The integral becomes $\int \frac{-du}{u} = -\ln|u| = -\ln|2 + \sin x|$.

Equating the results of the integrals and adding the constant of integration $C$:

$\ln|1+y| = -\ln|2 + \sin x| + C$

Rearrange the equation to combine the logarithmic terms:

$\ln|1+y| + \ln|2 + \sin x| = C$

$\ln(|(1+y)(2 + \sin x)|) = C$

Exponentiate both sides with base $e$:

$|(1+y)(2 + \sin x)| = e^C$

$(1+y)(2 + \sin x) = A$

Now, we use the given initial condition $y = 1$ when $x = 0$ to find the value of the constant $A$. Substitute $x=0$ and $y=1$ into equation (ii):

$(1+1)(2 + \sin 0) = A$

$(2)(2 + 0) = A$

$(2)(2) = A$

$A = 4$

Substitute the value of $A=4$ back into equation (ii) to obtain the particular solution:

Finally, we need to find the value of $y$ when $x = \frac{\pi}{2}$. Substitute $x = \frac{\pi}{2}$ into the particular solution (iii):

$\left(1+y\right)\left(2 + \sin \frac{\pi}{2}\right) = 4$

We know that $\sin \frac{\pi}{2} = 1$. Substitute this value:

$(1+y)(2 + 1) = 4$

$(1+y)(3) = 4$

Solve for $y$:

$1+y = \frac{4}{3}$

$y = \frac{4}{3} - 1$

$y = \frac{4}{3} - \frac{3}{3}$

$y = \frac{4 - 3}{3}$

$y = \frac{1}{3}$

Thus, the value of $y \left( \frac{\pi}{2} \right)$ is $\frac{1}{3}$.

Final Answer:

The value of $\mathbf{y \left( \frac{\pi}{2} \right)}$ is $\mathbf{\frac{1}{3}}$.

Given:

The differential equation $(1 + t) \frac{dy}{dt} – ty = 1$.

The initial condition $y(0) = – 1$.

To Show:

That $y(1) = -\frac{1}{2}$.

Solution:

The given differential equation is:

This is a linear first-order differential equation in $y$ with respect to $t$. To write it in the standard form $\frac{dy}{dt} + P(t)y = Q(t)$, we divide the entire equation by $(1+t)$, assuming $1+t \neq 0$ (i.e., $t \neq -1$).

$\frac{dy}{dt} - \frac{t}{1 + t} y = \frac{1}{1 + t}$

The integrating factor (IF) is given by $e^{\int P(t) dt}$.

Calculate the integral $\int P(t) dt$:

$\int P(t) dt = \int -\frac{t}{1 + t} dt$

We can rewrite the integrand as $\frac{-t}{1+t} = \frac{-(1+t) + 1}{1+t} = -1 + \frac{1}{1+t}$.

$\int \left(-1 + \frac{1}{1 + t}\right) dt = \int -1 \; dt + \int \frac{1}{1 + t} \; dt = -t + \ln|1+t|$

Calculate the integrating factor:

IF $= e^{\int P(t) dt} = e^{-t + \ln|1+t|} = e^{-t} \cdot e^{\ln|1+t|} = e^{-t} |1+t|$

For $t \ge 0$ (since our initial condition is at $t=0$ and we evaluate at $t=1$), $1+t \ge 1 > 0$, so $|1+t| = 1+t$.

The general solution of a linear first-order differential equation is given by:

$y \cdot (\text{IF}) = \int Q(t) \cdot (\text{IF}) \; dt + C$

Substitute $Q(t)$ and IF into the general solution formula:

$y \cdot \left( e^{-t} (1+t) \right) = \int \frac{1}{1 + t} \cdot \left( e^{-t} (1+t) \right) \; dt + C$

$y e^{-t} (1+t) = \int e^{-t} \; dt + C$

Evaluate the integral $\int e^{-t} \; dt$:

$\int e^{-t} \; dt = -e^{-t}$

Substitute the result of the integral back into the general solution equation:

Now, use the initial condition $y(0) = -1$ to find the value of the constant $C$. Substitute $t=0$ and $y=-1$ into equation (v):

$(-1) e^{-0} (1+0) = -e^{-0} + C$

$(-1) (1) (1) = -(1) + C$}

$-1 = -1 + C$

$C = 0$

Substitute the value of $C=0$ back into equation (v) to obtain the particular solution:

To find $y(1)$, substitute $t=1$ into equation (vi):

$y(1) e^{-1} (1+1) = -e^{-1}$}

$y(1) e^{-1} (2) = -e^{-1}$}

$2 y(1) e^{-1} = -e^{-1}$}

Divide both sides by $2 e^{-1}$ (since $e^{-1} \neq 0$):

$y(1) = \frac{-e^{-1}}{2 e^{-1}}$

$y(1) = -\frac{1}{2}$}

We have shown that $y(1) = -\frac{1}{2}$.

Given:

The general solution of the differential equation is $y = (\sin^{–1} x)^2 + A \cos^{–1} x + B$, where $A$ and $B$ are arbitrary constants.

To Form:

The differential equation having the given expression as its general solution.

Solution:

The given equation is:

Since there are two arbitrary constants ($A$ and $B$), the order of the required differential equation will be 2.

Differentiate equation (i) with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx} ((\sin^{-1} x)^2) + A \frac{d}{dx} (\cos^{-1} x) + \frac{d}{dx} (B)$

Using the chain rule and the standard derivatives $\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}$ and $\frac{d}{dx}(\cos^{-1} x) = -\frac{1}{\sqrt{1-x^2}}$, we get:

$\frac{dy}{dx} = 2 (\sin^{-1} x) \cdot \frac{1}{\sqrt{1-x^2}} + A \left(-\frac{1}{\sqrt{1-x^2}}\right) + 0$

$\frac{dy}{dx} = \frac{2 \sin^{-1} x}{\sqrt{1-x^2}} - \frac{A}{\sqrt{1-x^2}}$

Rearrange equation (ii) to isolate the term containing $\sin^{-1} x$ and the constant $A$:

Differentiate equation (iii) with respect to $x$ to eliminate the constant $A$. Use the product rule $\frac{d}{dx}(uv) = u'\nv + u\nv'$ on the left side:

Let $u = \sqrt{1-x^2}$ and $v = \frac{dy}{dx}$. Then $u' = \frac{1}{2\sqrt{1-x^2}} \cdot (-2x) = \frac{-x}{\sqrt{1-x^2}}$ and $v' = \frac{d^2y}{dx^2}$.

Left side differentiation: $\frac{d}{dx}\left(\sqrt{1-x^2} \frac{dy}{dx}\right) = \left(\frac{-x}{\sqrt{1-x^2}}\right) \frac{dy}{dx} + \sqrt{1-x^2} \frac{d^2y}{dx^2}$

Right side differentiation: $\frac{d}{dx}(2 \sin^{-1} x - A) = 2 \frac{d}{dx}(\sin^{-1} x) - \frac{d}{dx}(A) = 2 \cdot \frac{1}{\sqrt{1-x^2}} - 0 = \frac{2}{\sqrt{1-x^2}}$

Equating the derivatives of both sides:

$\frac{-x}{\sqrt{1-x^2}} \frac{dy}{dx} + \sqrt{1-x^2} \frac{d^2y}{dx^2} = \frac{2}{\sqrt{1-x^2}}$

Multiply the entire equation by $\sqrt{1-x^2}$ to clear the denominators:

$\sqrt{1-x^2} \left(\frac{-x}{\sqrt{1-x^2}} \frac{dy}{dx}\right) + \sqrt{1-x^2} \left(\sqrt{1-x^2} \frac{d^2y}{dx^2}\right) = \sqrt{1-x^2} \left(\frac{2}{\sqrt{1-x^2}}\right)$

$-x \frac{dy}{dx} + (1-x^2) \frac{d^2y}{dx^2} = 2$

Rearrange the terms to get the final differential equation:

$(1-x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = 2$

Final Answer:

The differential equation is $(1-x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = 2$.

Given:

Family of all circles which pass through the origin and whose centres lie on the y-axis.

To Form:

The differential equation representing this family of circles.

Solution:

The equation of a circle with centre at $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.

For the given family of circles, the centre lies on the y-axis, so $h=0$. The equation becomes:

$(x-0)^2 + (y-k)^2 = r^2$

Since the circle passes through the origin $(0,0)$, substitute $x=0$ and $y=0$ into equation (i):

$0^2 + (0-k)^2 = r^2$

$k^2 = r^2$

Substitute $r^2 = k^2$ back into equation (i):

$x^2 + (y-k)^2 = k^2$

Expand the equation:

$x^2 + y^2 - 2ky + k^2 = k^2$

$x^2 + y^2 - 2ky = 0$

Since there is one arbitrary constant $k$, the order of the differential equation will be 1. Differentiate equation (ii) with respect to $x$:

$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) - \frac{d}{dx}(2ky) = \frac{d}{dx}(0)$

$2x + 2y \frac{dy}{dx} - 2k \frac{dy}{dx} = 0$

Divide the entire equation by 2:

Now, we need to eliminate the constant $k$ from equations (ii) and (iii). From equation (ii), we can express $2k$ (assuming $y \neq 0$):

$2ky = x^2 + y^2$

$2k = \frac{x^2 + y^2}{y}$

Rearrange equation (iii) to isolate the term with $k$:

$k \frac{dy}{dx} = x + y \frac{dy}{dx}$

Multiply by 2 on both sides:

$2k \frac{dy}{dx} = 2x + 2y \frac{dy}{dx}$

Substitute the expression for $2k = \frac{x^2 + y^2}{y}$ into this equation:

$\left(\frac{x^2 + y^2}{y}\right) \frac{dy}{dx} = 2x + 2y \frac{dy}{dx}$

Multiply both sides by $y$ (assuming $y \neq 0$) to clear the denominator:

$(x^2 + y^2) \frac{dy}{dx} = y(2x + 2y \frac{dy}{dx})$

$(x^2 + y^2) \frac{dy}{dx} = 2xy + 2y^2 \frac{dy}{dx}$

Rearrange the terms to group $\frac{dy}{dx}$:

$(x^2 + y^2) \frac{dy}{dx} - 2y^2 \frac{dy}{dx} = 2xy$

$(x^2 + y^2 - 2y^2) \frac{dy}{dx} = 2xy$

$(x^2 - y^2) \frac{dy}{dx} = 2xy$

Note: The case where $y=0$ corresponds to the points $(0,0)$ and potentially other points on the x-axis. The original equation $x^2 + y^2 - 2ky = 0$ means that if $y=0$, then $x^2 = 0$, so $x=0$. Thus, only the origin $(0,0)$ is on the x-axis for these circles. At the origin, the concept of $\frac{dy}{dx}$ might be ambiguous for the family of circles. The derived differential equation $(x^2 - y^2) \frac{dy}{dx} = 2xy$ becomes $(0 - 0)\frac{dy}{dx} = 2(0)(0) \implies 0=0$ at the origin, which is consistent.

Final Answer:

The differential equation of all circles which pass through the origin and whose centres lie on the y-axis is $(x^2 - y^2) \frac{dy}{dx} = 2xy$.

Given:

The differential equation $(1 + x^2) \frac{dy}{dx} + 2xy = 4x^2$.

The initial condition is that the curve passes through the origin, i.e., $y(0) = 0$ (when $x=0$, $y=0$).

To Find:

The equation of the curve satisfying the given differential equation and initial condition.

Solution:

The given differential equation is:

$(1 + x^2) \frac{dy}{dx} + 2xy = 4x^2$

This is a linear first-order differential equation. To write it in the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we divide the entire equation by $(1 + x^2)$, assuming $1 + x^2 \neq 0$. Since $x^2 \ge 0$, $1+x^2$ is always positive, so this division is valid.

$\frac{dy}{dx} + \frac{2xy}{1 + x^2} = \frac{4x^2}{1 + x^2}$

Comparing equation (i) with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we have:

$P(x) = \frac{2x}{1 + x^2}$

$Q(x) = \frac{4x^2}{1 + x^2}$

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

Calculate the integral $\int P(x) dx$:

$\int P(x) dx = \int \frac{2x}{1 + x^2} dx$

Let $u = 1 + x^2$. Then $du = 2x dx$. The integral becomes:

$\int \frac{du}{u} = \ln|u| = \ln|1 + x^2|$

Since $1 + x^2 > 0$ for all real $x$, we have $\ln(1 + x^2)$.

Calculate the integrating factor:

IF $= e^{\int P(x) dx} = e^{\ln(1 + x^2)}$

IF $= 1 + x^2$

The general solution of a linear first-order differential equation is given by:

$y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) \; dx + C$

Substitute $Q(x)$ and IF into equation (ii):

$y(1 + x^2) = \int \frac{4x^2}{1 + x^2} \cdot (1 + x^2) \; dx + C$

$y(1 + x^2) = \int 4x^2 \; dx + C$

Evaluate the integral:

$\int 4x^2 \; dx = 4 \int x^2 \; dx = 4 \frac{x^{2+1}}{2+1} = 4 \frac{x^3}{3}$

Substitute the result of the integral back into the general solution equation:

This is the general solution of the differential equation.

Now, we use the initial condition $y(0) = 0$ to find the value of the constant $C$. Substitute $x=0$ and $y=0$ into equation (iii):

$0(1 + 0^2) = \frac{4(0)^3}{3} + C$

$0(1) = 0 + C$

$0 = C$

Substitute the value of $C=0$ back into the general solution (iii) to obtain the particular solution, which is the equation of the curve passing through the origin:

$y(1 + x^2) = \frac{4x^3}{3}$

We can write $y$ explicitly as a function of $x$:

$y = \frac{4x^3}{3(1 + x^2)}$

Final Answer:

The equation of the curve passing through the origin and satisfying the given differential equation is $y(1 + x^2) = \frac{4x^3}{3}$ or $y = \frac{4x^3}{3(1 + x^2)}$.

Given:

The differential equation $x^2 \frac{dy}{dx} = x^2 + xy + y^2$.

To Solve:

Find the general solution of the given differential equation.

Solution:

The given differential equation is:

$x^2 \frac{dy}{dx} = x^2 + xy + y^2$

Assuming $x \neq 0$, we can divide both sides by $x^2$ to write the equation in the form $\frac{dy}{dx} = f(x,y)$:

$\frac{dy}{dx} = \frac{x^2 + xy + y^2}{x^2}$

$\frac{dy}{dx} = \frac{x^2}{x^2} + \frac{xy}{x^2} + \frac{y^2}{x^2}$

This is a homogeneous differential equation because the right-hand side is a function of $\frac{y}{x}$.

We use the substitution $y = vx$, where $v$ is a function of $x$.

Differentiate $y = vx$ with respect to $x$ using the product rule:

Substitute $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into equation (i):

$v + x \frac{dv}{dx} = 1 + v + v^2$

Subtract $v$ from both sides:

$x \frac{dv}{dx} = 1 + v^2$

This is a variable separable differential equation. We can separate the variables $v$ and $x$ by dividing both sides by $(1+v^2)$ and multiplying by $dx$ (assuming $1+v^2 \neq 0$, which is always true for real $v$):

$\frac{dv}{1 + v^2} = \frac{dx}{x}$

Integrate both sides of the separated equation:

$\int \frac{dv}{1 + v^2} = \int \frac{dx}{x}$

Perform the integration:

$\tan^{-1} v = \ln|x| + C$

Substitute back $v = \frac{y}{x}$ into equation (iii):

$\tan^{-1} \left(\frac{y}{x}\right) = \ln|x| + C$

Final Answer:

The general solution of the differential equation is $\tan^{-1} \left(\frac{y}{x}\right) = \ln|x| + C$, where $\mathbf{C}$ is an arbitrary constant.

Given:

The differential equation $(1 + y^2) + (x – e^{\tan^{–1}y}) \frac{dy}{dx} = 0$.

To Solve:

Find the general solution of the given differential equation.

Solution:

The given differential equation is:

Rearrange the equation to isolate the $\frac{dy}{dx}$ term:

$(x – e^{\tan^{–1}y}) \frac{dy}{dx} = -(1 + y^2)$

Assuming $1+y^2 \neq 0$ (which is true for real $y$), we can write $\frac{dx}{dy}$ by taking the reciprocal of $\frac{dy}{dx}$ after rearranging:

$\frac{dx}{dy} = \frac{x – e^{\tan^{–1}y}}{-(1 + y^2)}$

$\frac{dx}{dy} = \frac{-(x – e^{\tan^{–1}y})}{1 + y^2}$

$\frac{dx}{dy} = \frac{e^{\tan^{–1}y} - x}{1 + y^2}$

Separate the terms on the right-hand side:

$\frac{dx}{dy} = \frac{e^{\tan^{–1}y}}{1 + y^2} - \frac{x}{1 + y^2}$

Rearrange the equation to write it in the standard form of a linear first-order differential equation in $x$ with respect to $y$: $\frac{dx}{dy} + P(y)x = Q(y)$

Comparing equation (ii) with the standard form $\frac{dx}{dy} + P(y)x = Q(y)$, we identify $P(y)$ and $Q(y)$:

$P(y) = \frac{1}{1 + y^2}$

$Q(y) = \frac{e^{\tan^{–1}y}}{1 + y^2}$

The integrating factor (IF) is given by $e^{\int P(y) dy}$.

Calculate the integral $\int P(y) dy$:

$\int P(y) dy = \int \frac{1}{1 + y^2} dy = \tan^{-1} y$

Calculate the integrating factor:

The general solution of a linear first-order differential equation in $x$ is given by:

$x \cdot (\text{IF}) = \int Q(y) \cdot (\text{IF}) \; dy + C$

Substitute $Q(y)$ and IF into equation (iv):

$x \cdot e^{\tan^{-1}y} = \int \frac{e^{\tan^{–1}y}}{1 + y^2} \cdot e^{\tan^{–1}y} \; dy + C$}

$x e^{\tan^{-1}y} = \int \frac{(e^{\tan^{-1}y})^2}{1 + y^2} \; dy + C$}

$x e^{\tan^{-1}y} = \int \frac{e^{2\tan^{-1}y}}{1 + y^2} \; dy + C$}

To evaluate the integral $\int \frac{e^{2\tan^{-1}y}}{1 + y^2} \; dy$, let $u = \tan^{-1} y$. Then $du = \frac{1}{1+y^2} dy$. The integral becomes:

$\int e^{2u} du = \frac{1}{2} e^{2u}$}

Substitute back $u = \tan^{-1} y$ into the integral result:

$\int \frac{e^{2\tan^{-1}y}}{1 + y^2} \; dy = \frac{1}{2} e^{2\tan^{-1}y}$}

Substitute the result of the integral back into the general solution equation:

$x e^{\tan^{-1}y} = \frac{1}{2} e^{2\tan^{-1}y} + C$}

We can express $x$ explicitly:

$x = \frac{\frac{1}{2} e^{2\tan^{-1}y} + C}{e^{\tan^{-1}y}}$

$x = \frac{1}{2} \frac{e^{2\tan^{-1}y}}{e^{\tan^{-1}y}} + \frac{C}{e^{\tan^{-1}y}}$

$x = \frac{1}{2} e^{\tan^{-1}y} + C e^{-\tan^{-1}y}$}

Final Answer:

The general solution of the differential equation is $x e^{\tan^{-1}y} = \frac{1}{2} e^{2\tan^{-1}y} + C$ or $x = \frac{1}{2} e^{\tan^{-1}y} + C e^{-\tan^{-1}y}$, where $\mathbf{C}$ is an arbitrary constant.

Given:

The differential equation $y^2 dx + (x^2 – xy + y^2) dy = 0$.

To Solve:

Find the general solution of the given differential equation.

Solution:

The given differential equation is:

$y^2 dx + (x^2 – xy + y^2) dy = 0$

We can rewrite the equation in the form $\frac{dx}{dy} = F(x,y)$ (assuming $dy \neq 0$):

$y^2 dx = -(x^2 – xy + y^2) dy$

$\frac{dx}{dy} = - \frac{x^2 – xy + y^2}{y^2}$

$\frac{dx}{dy} = - \left( \frac{x^2}{y^2} - \frac{xy}{y^2} + \frac{y^2}{y^2} \right)$

This is a homogeneous differential equation because the right-hand side can be expressed as a function of $\frac{x}{y}$.

We use the substitution $x = vy$, where $v$ is a function of $y$.

Differentiate $x = vy$ with respect to $y$ using the product rule:

Substitute $x = vy$ (so $\frac{x}{y} = v$) and $\frac{dx}{dy} = v + y \frac{dv}{dy}$ into equation (i):

$v + y \frac{dv}{dy} = - (v^2 - v + 1)$

$v + y \frac{dv}{dy} = -v^2 + v - 1$}

Subtract $v$ from both sides:

$y \frac{dv}{dy} = -v^2 + v - 1 - v$}

$y \frac{dv}{dy} = -v^2 - 1$}

$y \frac{dv}{dy} = -(v^2 + 1)$

This is a variable separable differential equation. We can separate the variables $v$ and $y$ by dividing both sides by $(v^2 + 1)$ and multiplying by $dy$ (assuming $v^2+1 \neq 0$, which is always true for real $v$, and assuming $y \neq 0$):

$\frac{dv}{v^2 + 1} = - \frac{dy}{y}$

Integrate both sides of the separated equation:

$\int \frac{dv}{v^2 + 1} = \int - \frac{dy}{y}$

Perform the integration:

$\int \frac{dv}{v^2 + 1} = \tan^{-1} v$}

$\int - \frac{dy}{y} = -\int \frac{dy}{y} = -\ln|y| + C'$

Equating the results of the integrals:

Substitute back $v = \frac{x}{y}$ into equation (iii):

We can rearrange this solution as:

$\tan^{-1} \left(\frac{x}{y}\right) + \ln|y| = C$

Note: The case $y=0$ yields $x^2 dy = 0$ from the original equation. If $x \neq 0$, then $dy=0$, meaning $y$ is constant (and equal to 0). Thus $y=0$ is a solution. The derived general solution $\tan^{-1} \left(\frac{x}{y}\right) + \ln|y| = C$ requires $y \neq 0$ and does not include the singular solution $y=0$. However, the standard general solution obtained through this method is considered complete for $y \neq 0$.

Final Answer:

The general solution of the differential equation is $\tan^{-1} \left(\frac{x}{y}\right) + \ln|y| = C$, where $\mathbf{C}$ is an arbitrary constant.

Given:

The differential equation $(x + y) (dx – dy) = dx + dy$.

To Solve:

Find the general solution of the given differential equation.

Solution:

The given differential equation is:

Expand the left side of the equation:

$(x+y)dx - (x+y)dy = dx + dy$

Rearrange the terms to group $dx$ and $dy$:

$(x+y)dx - dx = dy + (x+y)dy$

$(x+y-1)dx = (1+x+y)dy$

Assuming $dx \neq 0$ and $1+x+y \neq 0$, we can write the equation in the form $\frac{dy}{dx}$:

We use the substitution suggested in the hint: $z = x+y$.

Differentiate this substitution with respect to $x$:

$\frac{dz}{dx} = \frac{d}{dx}(x+y)$

$\frac{dz}{dx} = 1 + \frac{dy}{dx}$

From this, we can express $\frac{dy}{dx}$ as:

Substitute equation (iii) and $z = x+y$ into equation (ii):

$\frac{dz}{dx} - 1 = \frac{z-1}{z+1}$

Add 1 to both sides:

$\frac{dz}{dx} = 1 + \frac{z-1}{z+1}$

$\frac{dz}{dx} = \frac{z+1}{z+1} + \frac{z-1}{z+1}$

$\frac{dz}{dx} = \frac{(z+1) + (z-1)}{z+1}$

Equation (iv) is a variable separable differential equation. Separate the variables $z$ and $x$, assuming $z \neq 0$:

$\frac{z+1}{2z} dz = dx$

$\frac{1}{2} \left( \frac{z}{z} + \frac{1}{z} \right) dz = dx$

$\frac{1}{2} \left( 1 + \frac{1}{z} \right) dz = dx$

Integrate both sides of the separated equation:

$\int \frac{1}{2} \left( 1 + \frac{1}{z} \right) dz = \int dx$

$\frac{1}{2} \int \left( 1 + \frac{1}{z} \right) dz = \int dx$

$\frac{1}{2} (z + \ln|z|) = x + C'$

Multiply the equation by 2:

$z + \ln|z| = 2x + 2C'$

Let $C = 2C'$, where $C$ is a new arbitrary constant.

Substitute back $z = x+y$ into equation (v):

$(x+y) + \ln|x+y| = 2x + C$

Rearrange the terms to get the general solution:

$\ln|x+y| = 2x - (x+y) + C$

$\ln|x+y| = 2x - x - y + C$

$\ln|x+y| = x - y + C$

Final Answer:

The general solution of the differential equation is $\ln|x+y| = x - y + C$, where $\mathbf{C}$ is an arbitrary constant.

Given:

The differential equation $2 (y + 3) – xy \frac{dy}{dx} = 0$.

The initial condition $y(1) = -2$, which means $y = -2$ when $x = 1$.

To Solve:

Find the particular solution of the given differential equation that satisfies the initial condition.

Solution:

The given differential equation is:

$2 (y + 3) – xy \frac{dy}{dx} = 0$}

Rearrange the equation to separate the terms and write it in a suitable form for separation of variables:

$xy \frac{dy}{dx} = 2 (y + 3)$

Assuming $x \neq 0$ and $y+3 \neq 0$, we can separate the variables by dividing both sides by $x(y+3)$ and multiplying by $dx$:

$\frac{y}{y + 3} dy = \frac{2}{x} dx$

Integrate both sides of the separated equation:

$\int \frac{y}{y + 3} dy = \int \frac{2}{x} dx$}

For the integral on the left-hand side, we rewrite the numerator:

$\int \frac{y}{y + 3} dy = \int \frac{y+3-3}{y + 3} dy = \int \left(1 - \frac{3}{y+3}\right) dy$

Evaluate the integral:

$\int \left(1 - \frac{3}{y+3}\right) dy = \int 1 \, dy - \int \frac{3}{y+3} dy = y - 3 \log|y+3|$

For the integral on the right-hand side:

$\int \frac{2}{x} dx = 2 \int \frac{1}{x} dx = 2 \log|x|$

Equating the results of the integrals and adding the arbitrary constant of integration $C$:

This is the general solution of the differential equation.

Now, we use the given initial condition $y = -2$ when $x = 1$ to find the value of the constant $C$. Substitute $x=1$ and $y=-2$ into equation (i):

$-2 - 3 \log|-2+3| = 2 \log|1| + C$}

$-2 - 3 \log(1) = 2 \log(1) + C$}

Since $\log(1) = 0$:

$-2 - 3(0) = 2(0) + C$}

$-2 - 0 = 0 + C$}

Substitute the value of $C = -2$ back into the general solution (i) to obtain the particular solution that satisfies the initial condition:

This is the equation of the curve passing through the point $(1, -2)$ and satisfying the differential equation.

Final Answer:

The particular solution of the differential equation is $y - 3 \log|y+3| = 2 \log|x| - 2$.

Given:

The differential equation $dy = \cos x (2 – y \text{ cosec } x) dx$.

The initial condition $y = 2$ when $x = \frac{\pi}{2}$.

To Solve:

Find the particular solution of the given differential equation that satisfies the initial condition.

Solution:

The given differential equation is:

$dy = \cos x (2 – y \text{ cosec } x) dx$

Divide both sides by $dx$ (assuming $dx \neq 0$):

$\frac{dy}{dx} = \cos x (2 – y \text{ cosec } x)$

Expand the right-hand side:

$\frac{dy}{dx} = 2 \cos x – y \cos x \text{ cosec } x$

Recall that $\text{cosec } x = \frac{1}{\sin x}$. Substitute this into the equation:

$\frac{dy}{dx} = 2 \cos x – y \cos x \cdot \frac{1}{\sin x}$

$\frac{dy}{dx} = 2 \cos x – y \frac{\cos x}{\sin x}$

$\frac{dy}{dx} = 2 \cos x – y \cot x$

Rearrange the equation to write it in the standard form of a linear first-order differential equation $\frac{dy}{dx} + P(x)y = Q(x)$:

Comparing equation (i) with the standard form, we identify $P(x)$ and $Q(x)$:

$P(x) = \cot x$

$Q(x) = 2 \cos x$}

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

Calculate the integral $\int P(x) dx$:

$\int P(x) dx = \int \cot x dx = \int \frac{\cos x}{\sin x} dx$}

Let $u = \sin x$, then $du = \cos x dx$. The integral becomes $\int \frac{du}{u} = \log|u| = \log|\sin x|$.

Calculate the integrating factor:

IF $= e^{\int P(x) dx} = e^{\log|\sin x|} = |\sin x|$.

Since the initial condition is given at $x = \frac{\pi}{2}$, and $\sin x > 0$ near $x = \frac{\pi}{2}$, we can take IF $= \sin x$.}

The general solution of a linear first-order differential equation is given by:

$y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) \; dx + C$

Substitute $Q(x)$ and IF into equation (iii):

$y \cdot \sin x = \int (2 \cos x) \cdot (\sin x) \; dx + C$}

$y \sin x = \int 2 \sin x \cos x \; dx + C$}

Use the identity $\sin(2x) = 2 \sin x \cos x$:

$y \sin x = \int \sin(2x) \; dx + C$}

Evaluate the integral $\int \sin(2x) \; dx$:

$\int \sin(2x) \; dx = -\frac{\cos(2x)}{2}$}

Substitute the result of the integral back into the general solution equation:

This is the general solution of the differential equation.

Now, we use the given initial condition $y = 2$ when $x = \frac{\pi}{2}$ to find the value of the constant $C$. Substitute $x=\frac{\pi}{2}$ and $y=2$ into equation (iv):

$2 \sin\left(\frac{\pi}{2}\right) = -\frac{\cos\left(2 \cdot \frac{\pi}{2}\right)}{2} + C$}

$2 \sin\left(\frac{\pi}{2}\right) = -\frac{\cos(\pi)}{2} + C$}

We know that $\sin\left(\frac{\pi}{2}\right) = 1$ and $\cos(\pi) = -1$. Substitute these values:

$2(1) = -\frac{(-1)}{2} + C$}

$2 = \frac{1}{2} + C$}

Solve for $C$:

$C = 2 - \frac{1}{2}$}

$C = \frac{4}{2} - \frac{1}{2}$}

Substitute the value of $C = \frac{3}{2}$ back into the general solution (iv) to obtain the particular solution that satisfies the initial condition:

We can multiply the equation by 2 to eliminate fractions:

$2y \sin x = -\cos(2x) + 3$}

Final Answer:

The particular solution of the differential equation is $y \sin x = -\frac{\cos(2x)}{2} + \frac{3}{2}$ or $2y \sin x = 3 - \cos(2x)$.

Given:

The equation of the family of curves is $Ax^2 + By^2 = 1$, where $A$ and $B$ are arbitrary constants.

To Form:

The differential equation by eliminating the arbitrary constants $A$ and $B$.

Solution:

The given equation is:

Since there are two arbitrary constants ($A$ and $B$), we expect a second-order differential equation.

Differentiate equation (i) with respect to $x$:

$\frac{d}{dx}(Ax^2) + \frac{d}{dx}(By^2) = \frac{d}{dx}(1)$

$A(2x) + B(2y) \frac{dy}{dx} = 0$

Divide by 2:

Differentiate equation (ii) with respect to $x$ using the product rule:

$\frac{d}{dx}(Ax) + \frac{d}{dx}\left(By \frac{dy}{dx}\right) = \frac{d}{dx}(0)$

$A(1) + B \left[ \frac{d}{dx}(y) \frac{dy}{dx} + y \frac{d}{dx}\left(\frac{dy}{dx}\right) \right] = 0$

$A + B \left[ \left(\frac{dy}{dx}\right) \left(\frac{dy}{dx}\right) + y \frac{d^2y}{dx^2} \right] = 0$

Now we need to eliminate the constants $A$ and $B$ using equations (ii) and (iii).

From equation (ii), assuming $x \neq 0$, we can express $A$:

From equation (iii), assuming $B \neq 0$, we can express $A$:

Substitute the expression for $A$ from (v) into (iv):

$\left(-B \left[ \left(\frac{dy}{dx}\right)^2 + y \frac{d^2y}{dx^2} \right]\right) x = -By \frac{dy}{dx}$

Assuming $B \neq 0$, we can divide both sides by $-B$:

$\left[ \left(\frac{dy}{dx}\right)^2 + y \frac{d^2y}{dx^2} \right] x = y \frac{dy}{dx}$

Expand the left side:

$x \left(\frac{dy}{dx}\right)^2 + xy \frac{d^2y}{dx^2} = y \frac{dy}{dx}$

Rearrange the terms to write the differential equation:

$xy \frac{d^2y}{dx^2} + x \left(\frac{dy}{dx}\right)^2 - y \frac{dy}{dx} = 0$

Final Answer:

The differential equation is $xy \frac{d^2y}{dx^2} + x \left(\frac{dy}{dx}\right)^2 - y \frac{dy}{dx} = 0$.

Given:

The differential equation $(1 + y^2) \tan^{–1} x dx + 2y (1 + x^2) dy = 0$.

To Solve:

Find the general solution of the given differential equation.

Solution:

The given differential equation is:

$(1 + y^2) \tan^{-1} x dx + 2y (1 + x^2) dy = 0$}

This is a variable separable differential equation. We can separate the terms involving $x$ and $dx$ from the terms involving $y$ and $dy$ by moving one term to the other side and then dividing:

$(1 + y^2) \tan^{-1} x dx = -2y (1 + x^2) dy$

Divide both sides by $(1 + y^2)(1 + x^2)$ (assuming $1+y^2 \neq 0$ and $1+x^2 \neq 0$, which is true for real $x$ and $y$) and by $\tan^{-1} x$ (assuming $\tan^{-1} x \neq 0$, i.e., $x \neq 0$):

$\frac{(1 + y^2) \tan^{-1} x}{(1 + y^2)(1 + x^2)} dx = - \frac{2y (1 + x^2)}{(1 + y^2)(1 + x^2)} dy$

$\frac{\tan^{-1} x}{1 + x^2} dx = - \frac{2y}{1 + y^2} dy$

Integrate both sides of the separated equation:

$\int \frac{\tan^{-1} x}{1 + x^2} dx = \int - \frac{2y}{1 + y^2} dy$

For the integral on the left-hand side, let $u = \tan^{-1} x$. Then $du = \frac{1}{1+x^2} dx$. The integral becomes $\int u \, du = \frac{u^2}{2}$. Substitute back $u = \tan^{-1} x$:

$\int \frac{\tan^{-1} x}{1 + x^2} dx = \frac{(\tan^{-1} x)^2}{2}$}

For the integral on the right-hand side, let $v = 1 + y^2$. Then $dv = 2y \, dy$. The integral becomes $\int - \frac{dv}{v} = -\log|v|$. Substitute back $v = 1 + y^2$:

$\int - \frac{2y}{1 + y^2} dy = -\log|1 + y^2|$

Since $1+y^2 > 0$ for real $y$, we have $-\log(1 + y^2)$.

Equating the results of the integrals and adding the arbitrary constant of integration $C$:

$\frac{(\tan^{-1} x)^2}{2} = -\log(1 + y^2) + C$

Rearrange the terms to write the general solution:

$\frac{1}{2} (\tan^{-1} x)^2 + \log(1 + y^2) = C$

We can multiply by 2 to eliminate the fraction:

$(\tan^{-1} x)^2 + 2 \log(1 + y^2) = 2C$

Let $K = 2C$, where $K$ is a new arbitrary constant.

Note: The separation step assumed $\tan^{-1} x \neq 0$, i.e., $x \neq 0$. If $x=0$, the original equation becomes $(1+y^2) \tan^{-1}(0) dx + 2y(1+0^2) dy = 0$, which simplifies to $0 + 2y dy = 0$, or $2y dy = 0$. This implies $y=0$ or $dy=0$. The solution $y=C'$ is obtained by integrating $dy=0$. If $y=0$, we get $0=0$. The solution $y=0$ corresponds to $y=C'$ with $C'=0$. Let's check if $y=0$ is included in the general solution $(\tan^{-1} x)^2 + 2 \log(1 + y^2) = K$. If $y=0$, we get $(\tan^{-1} x)^2 + 2 \log(1) = K$, so $(\tan^{-1} x)^2 = K$. This is not a trivial solution. However, the variable separable method is generally applied where the denominators are non-zero. The solution obtained covers the regions where the separation is valid.

Final Answer:

The general solution of the differential equation is $(\tan^{-1} x)^2 + 2 \log(1 + y^2) = K$, where $\mathbf{K}$ is an arbitrary constant.

Given:

Family of concentric circles with centre at $(1, 2)$.

To Find:

The differential equation representing this family of circles.

Solution:

The equation of a circle with centre at $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.

For the given family of circles, the centre is fixed at $(1, 2)$, so $h=1$ and $k=2$. The radius $r$ is the only parameter that varies, representing the different circles in the concentric system. The equation of the family of circles is:

Since there is only one arbitrary constant ($r^2$ or $r$), the order of the differential equation will be 1.

Differentiate equation (i) with respect to $x$:

$\frac{d}{dx}((x-1)^2) + \frac{d}{dx}((y-2)^2) = \frac{d}{dx}(r^2)$

Using the chain rule, we get:

$2(x-1) \cdot \frac{d}{dx}(x-1) + 2(y-2) \cdot \frac{d}{dx}(y-2) = 0$

$2(x-1)(1) + 2(y-2) \left(\frac{dy}{dx} - 0\right) = 0$

$2(x-1) + 2(y-2) \frac{dy}{dx} = 0$

Divide the entire equation by 2:

$(x-1) + (y-2) \frac{dy}{dx} = 0$

This differential equation does not contain the arbitrary constant $r$.

We can also express $\frac{dy}{dx}$ explicitly:

$(y-2) \frac{dy}{dx} = -(x-1)$

$\frac{dy}{dx} = - \frac{x-1}{y-2}$

$\frac{dy}{dx} = \frac{1-x}{y-2}$

Final Answer:

The differential equation of the system of concentric circles with centre $(1, 2)$ is $(x-1) + (y-2) \frac{dy}{dx} = 0$ or $\frac{dy}{dx} = \frac{1-x}{y-2}$.

Given:

The differential equation $y + \frac{d}{dx} (xy) = x (\sin x + \log x)$.

To Solve:

Find the general solution of the given differential equation.

Solution:

The given differential equation is:

First, we evaluate the derivative term $\frac{d}{dx} (xy)$ using the product rule:

$\frac{d}{dx} (xy) = \frac{d}{dx}(x) \cdot y + x \cdot \frac{d}{dx}(y) = 1 \cdot y + x \frac{dy}{dx} = y + x \frac{dy}{dx}$

Substitute this result back into the original differential equation (i):

$y + \left(y + x \frac{dy}{dx}\right) = x (\sin x + \log x)$

$2y + x \frac{dy}{dx} = x \sin x + x \log x$

Rearrange the equation to write it in the standard form of a linear first-order differential equation $\frac{dy}{dx} + P(x)y = Q(x)$. Divide the entire equation by $x$ (assuming $x \neq 0$):

$\frac{dy}{dx} + \frac{2y}{x} = \frac{x \sin x + x \log x}{x}$

Comparing equation (ii) with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we identify $P(x)$ and $Q(x)$:

$P(x) = \frac{2}{x}$

$Q(x) = \sin x + \log x$

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

Calculate the integral $\int P(x) dx$:

$\int P(x) dx = \int \frac{2}{x} dx = 2 \int \frac{1}{x} dx = 2 \log|x|$

Assuming $x > 0$ (since $\log x$ is present in the original equation), we have $\int P(x) dx = 2 \log x = \log(x^2)$.

Calculate the integrating factor:

The general solution of a linear first-order differential equation is given by:

$y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) \; dx + C$

Substitute $Q(x)$ and IF into equation (iv):

$y \cdot x^2 = \int (\sin x + \log x) \cdot x^2 \; dx + C$

$y x^2 = \int (x^2 \sin x + x^2 \log x) \; dx + C$

$y x^2 = \int x^2 \sin x \; dx + \int x^2 \log x \; dx + C$

Evaluate the integrals using integration by parts.

Integral 1: $\int x^2 \sin x \; dx$

Using $\int u \; dv = uv - \int v \; du$ twice:

$\int x^2 \sin x \; dx = -x^2 \cos x - \int (-\cos x)(2x) dx = -x^2 \cos x + 2 \int x \cos x dx$

$\int x \cos x dx = x \sin x - \int \sin x dx = x \sin x - (-\cos x) = x \sin x + \cos x$

So, $\int x^2 \sin x \; dx = -x^2 \cos x + 2(x \sin x + \cos x) = -x^2 \cos x + 2x \sin x + 2 \cos x = (2-x^2)\cos x + 2x \sin x$

Integral 2: $\int x^2 \log x \; dx$

Using $\int u \; dv = uv - \int v \; du$ with $u = \log x$ and $dv = x^2 dx$:

$\int x^2 \log x \; dx = (\log x) \left(\frac{x^3}{3}\right) - \int \left(\frac{x^3}{3}\right) \left(\frac{1}{x}\right) dx = \frac{x^3}{3} \log x - \int \frac{x^2}{3} dx$

$= \frac{x^3}{3} \log x - \frac{1}{3} \left(\frac{x^3}{3}\right) = \frac{x^3}{3} \log x - \frac{x^3}{9}$

Substitute the results of both integrals back into the general solution equation $y x^2 = \int x^2 \sin x \; dx + \int x^2 \log x \; dx + C$:

$y x^2 = (2-x^2)\cos x + 2x \sin x + \frac{x^3}{3} \log x - \frac{x^3}{9} + C$

Divide by $x^2$ to express $y$ explicitly:

$y = \frac{(2-x^2)\cos x + 2x \sin x}{x^2} + \frac{\frac{x^3}{3} \log x - \frac{x^3}{9}}{x^2} + \frac{C}{x^2}$

$y = \frac{2-x^2}{x^2}\cos x + \frac{2x}{x^2}\sin x + \frac{x^3}{3x^2}\log x - \frac{x^3}{9x^2} + \frac{C}{x^2}$

Final Answer:

The general solution of the differential equation is $y x^2 = (2-x^2)\cos x + 2x \sin x + \frac{x^3}{3} \log x - \frac{x^3}{9} + C$ or $y = \left(\frac{2}{x^2} - 1\right) \cos x + \frac{2}{x} \sin x + \frac{x}{3} \log x - \frac{x}{9} + \frac{C}{x^2}$, where $\mathbf{C}$ is an arbitrary constant.

Given:

The differential equation $(1 + \tan y) (dx – dy) + 2xdy = 0$.

To Solve:

Find the general solution of the given differential equation.

Solution:

The given differential equation is:

Expand the term $(1 + \tan y) (dx – dy)$:

$(1 + \tan y) dx - (1 + \tan y) dy + 2xdy = 0$}

Group the $dx$ and $dy$ terms:

$(1 + \tan y) dx + (2x - (1 + \tan y)) dy = 0$}

$(1 + \tan y) dx + (2x - 1 - \tan y) dy = 0$}

Assuming $dy \neq 0$, we can write the equation in the form $\frac{dx}{dy}$:

$(1 + \tan y) \frac{dx}{dy} + (2x - 1 - \tan y) = 0$}

$(1 + \tan y) \frac{dx}{dy} = -(2x - 1 - \tan y)$}

$(1 + \tan y) \frac{dx}{dy} = -2x + 1 + \tan y$}

Divide by $(1 + \tan y)$ (assuming $1+\tan y \neq 0$):

$\frac{dx}{dy} = \frac{-2x + 1 + \tan y}{1 + \tan y}$}

$\frac{dx}{dy} = \frac{-2x}{1 + \tan y} + \frac{1 + \tan y}{1 + \tan y}$}

$\frac{dx}{dy} = - \frac{2}{1 + \tan y} x + 1$}

Rearrange the equation to write it in the standard form of a linear first-order differential equation in $x$ with respect to $y$: $\frac{dx}{dy} + P(y)x = Q(y)$

Comparing equation (ii) with the standard form, we identify $P(y)$ and $Q(y)$:

$P(y) = \frac{2}{1 + \tan y}$}

$Q(y) = 1$}

The integrating factor (IF) is given by $e^{\int P(y) dy}$.

Calculate the integral $\int P(y) dy$:

$\int P(y) dy = \int \frac{2}{1 + \tan y} dy = \int \frac{2}{1 + \frac{\sin y}{\cos y}} dy = \int \frac{2 \cos y}{\cos y + \sin y} dy$}

Let's use a trick for integrals of the form $\int \frac{a \cos x + b \sin x}{c \cos x + d \sin x} dx$. Here, we want $\int \frac{2 \cos y}{\cos y + \sin y} dy$. We can write the numerator $2 \cos y$ as a linear combination of the denominator and its derivative.

Let $2 \cos y = A (\cos y + \sin y) + B (-\sin y + \cos y)$

$2 \cos y = (A+B) \cos y + (A-B) \sin y$}

Comparing coefficients of $\cos y$ and $\sin y$:

$A+B = 2$}

$A-B = 0$}

Adding the two equations: $2A = 2 \implies A = 1$. Subtracting the second from the first: $2B = 2 \implies B = 1$.

So, $2 \cos y = 1 (\cos y + \sin y) + 1 (-\sin y + \cos y)$.

The integral becomes:

$\int \frac{(\cos y + \sin y) + (-\sin y + \cos y)}{\cos y + \sin y} dy = \int \left(1 + \frac{-\sin y + \cos y}{\cos y + \sin y}\right) dy$}

$= \int 1 \; dy + \int \frac{\frac{d}{dy}(\cos y + \sin y)}{\cos y + \sin y} dy$}

$= y + \log|\cos y + \sin y|$

Calculate the integrating factor:

For simplicity, we can use IF $= e^y (\cos y + \sin y)$, absorbing the absolute value into the constant later.

The general solution of a linear first-order differential equation in $x$ is given by:

$x \cdot (\text{IF}) = \int Q(y) \cdot (\text{IF}) \; dy + C$

Substitute $Q(y)$ and IF into equation (iv):

$x \cdot e^y (\cos y + \sin y) = \int 1 \cdot e^y (\cos y + \sin y) \; dy + C$}

$x e^y (\cos y + \sin y) = \int e^y (\cos y + \sin y) \; dy + C$}

Evaluate the integral $\int e^y (\cos y + \sin y) \; dy$. This integral is of the form $\int e^{ay} (f(y) + f'(y)) dy$, where $f(y) = \sin y$ and $a=1$. The derivative of $f(y) = \sin y$ is $f'(y) = \cos y$. The formula for this type of integral is $\int e^{ay} (f(y) + f'(y)) dy = e^{ay} f(y) + C_{int}$.

Here, $\int e^y (\cos y + \sin y) \; dy = e^y \sin y$

Substitute the result of the integral back into the general solution equation:

$x e^y (\cos y + \sin y) = e^y \sin y + C$}

Final Answer:

The general solution of the differential equation is $x e^y (\cos y + \sin y) = e^y \sin y + C$, where $\mathbf{C}$ is an arbitrary constant.

Given:

The differential equation $\frac{dy}{dx} = \cos(x + y) + \sin (x + y)$.

To Solve:

Find the general solution of the given differential equation.

Solution:

The given differential equation is:

We use the substitution suggested: $z = x + y$.

Differentiate the substitution with respect to $x$:

$\frac{dz}{dx} = \frac{d}{dx}(x + y)$

$\frac{dz}{dx} = 1 + \frac{dy}{dx}$

From this, we can express $\frac{dy}{dx}$:

Substitute equation (ii) and $z = x+y$ into equation (i):

$\frac{dz}{dx} - 1 = \cos(z) + \sin(z)$

Add 1 to both sides:

Equation (iii) is a variable separable differential equation. Separate the variables $z$ and $x$:

$\frac{dz}{1 + \cos z + \sin z} = dx$

Integrate both sides:

$\int \frac{dz}{1 + \cos z + \sin z} = \int dx$

Evaluate the integral on the right side:

$\int dx = x + C'$

Evaluate the integral on the left side, $\int \frac{dz}{1 + \cos z + \sin z}$. We can use the substitution $t = \tan\left(\frac{z}{2}\right)$. Then $\cos z = \frac{1-t^2}{1+t^2}$, $\sin z = \frac{2t}{1+t^2}$, and $dz = \frac{2 dt}{1+t^2}$.

$1 + \cos z + \sin z = 1 + \frac{1-t^2}{1+t^2} + \frac{2t}{1+t^2} = \frac{1+t^2 + 1-t^2 + 2t}{1+t^2} = \frac{2 + 2t}{1+t^2} = \frac{2(1+t)}{1+t^2}$

The integral becomes:

$\int \frac{1}{\frac{2(1+t)}{1+t^2}} \cdot \frac{2 dt}{1+t^2} = \int \frac{1+t^2}{2(1+t)} \cdot \frac{2 dt}{1+t^2} = \int \frac{1}{1+t} dt$

$\int \frac{1}{1+t} dt = \log|1+t|$

Substitute back $t = \tan\left(\frac{z}{2}\right)$:

$\int \frac{dz}{1 + \cos z + \sin z} = \log\left|1 + \tan\left(\frac{z}{2}\right)\right|$

Equating the results of the integrals:

$\log\left|1 + \tan\left(\frac{z}{2}\right)\right| = x + C'$

Substitute back $z = x+y$:

Final Answer:

The general solution of the differential equation is $\log\left|1 + \tan\left(\frac{x+y}{2}\right)\right| = x + C$, where $\mathbf{C}$ is an arbitrary constant.

Given:

The differential equation $\frac{dy}{dx} - 3y = \sin 2x$.

To Find:

The general solution of the given differential equation.

Solution:

The given differential equation is:

This is a linear first-order differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.

Comparing equation (i) with the standard form, we identify $P(x)$ and $Q(x)$:

$P(x) = -3$

$Q(x) = \sin 2x$

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

Calculate the integral $\int P(x) dx$:

$\int P(x) dx = \int -3 dx = -3x$

Calculate the integrating factor:

The general solution of a linear first-order differential equation is given by:

$y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) \; dx + C$

Substitute $Q(x)$ and IF into the general solution formula (iii):

$y \cdot e^{-3x} = \int (\sin 2x) \cdot e^{-3x} \; dx + C$

$y e^{-3x} = \int e^{-3x} \sin 2x \; dx + C$

Evaluate the integral $\int e^{-3x} \sin 2x \; dx$. This integral can be solved using integration by parts or by using the formula $\int e^{ax} \sin(bx) dx = \frac{e^{ax}}{a^2+b^2}(a \sin(bx) - b \cos(bx))$.

Using the formula with $a = -3$ and $b = 2$:

$\int e^{-3x} \sin 2x \; dx = \frac{e^{-3x}}{(-3)^2 + 2^2}((-3) \sin 2x - 2 \cos 2x)$

$= \frac{e^{-3x}}{9 + 4}(-3 \sin 2x - 2 \cos 2x)$

$= \frac{e^{-3x}}{13}(-3 \sin 2x - 2 \cos 2x)$

Substitute the result of the integral back into the general solution equation:

Divide both sides of equation (iv) by $e^{-3x}$ to solve for $y$:

$y = \frac{\frac{e^{-3x}}{13}(-3 \sin 2x - 2 \cos 2x)}{e^{-3x}} + \frac{C}{e^{-3x}}$

$y = \frac{1}{13}(-3 \sin 2x - 2 \cos 2x) + C e^{3x}$

$y = - \frac{1}{13}(3 \sin 2x + 2 \cos 2x) + C e^{3x}$

Final Answer:

The general solution of the differential equation is $y = C e^{3x} - \frac{1}{13}(3 \sin 2x + 2 \cos 2x)$, where $\mathbf{C}$ is an arbitrary constant.

Given:

The slope of the tangent to the curve at any point $(x, y)$ is $\frac{x^2 + y^2}{2xy}$.

The curve passes through the point $(2, 1)$.

To Find:

The equation of the curve.

Solution:

The slope of the tangent to the curve at any point $(x, y)$ is given by $\frac{dy}{dx}$. So, the differential equation of the curve is:

We can check if this is a homogeneous differential equation by writing the right-hand side as a function of $\frac{y}{x}$.

$\frac{x^2 + y^2}{2xy} = \frac{\frac{x^2 + y^2}{x^2}}{\frac{2xy}{x^2}} = \frac{1 + \frac{y^2}{x^2}}{\frac{2y}{x}} = \frac{1 + \left(\frac{y}{x}\right)^2}{2\left(\frac{y}{x}\right)}$

Since the right-hand side is a function of $\frac{y}{x}$, it is a homogeneous differential equation.

We use the substitution $y = vx$, where $v$ is a function of $x$.

Differentiate $y = vx$ with respect to $x$ using the product rule:

Substitute equation (ii) and $y = vx$ (so $\frac{y}{x} = v$) into equation (i):

$v + x \frac{dv}{dx} = \frac{1 + v^2}{2v}$

Subtract $v$ from both sides:

$x \frac{dv}{dx} = \frac{1 + v^2}{2v} - v$

$x \frac{dv}{dx} = \frac{1 + v^2}{2v} - \frac{2v^2}{2v}$

$x \frac{dv}{dx} = \frac{1 + v^2 - 2v^2}{2v}$

$x \frac{dv}{dx} = \frac{1 - v^2}{2v}$

This is a variable separable differential equation. Separate the variables $v$ and $x$ by dividing both sides by $\frac{1-v^2}{2v}$ and multiplying by $dx$ (assuming $1-v^2 \neq 0$, i.e., $v \neq \pm 1$, and $x \neq 0$):

$\frac{2v}{1 - v^2} dv = \frac{dx}{x}$

Integrate both sides of the separated equation:

$\int \frac{2v}{1 - v^2} dv = \int \frac{dx}{x}$

For the integral on the left-hand side, let $u = 1 - v^2$. Then $du = -2v \, dv$. So $2v \, dv = -du$. The integral becomes $\int \frac{-du}{u} = -\log|u|$. Substitute back $u = 1 - v^2$:

$\int \frac{2v}{1 - v^2} dv = -\log|1 - v^2|$

For the integral on the right-hand side:

$\int \frac{dx}{x} = \log|x| + C'$

Equating the results of the integrals:

$-\log|1 - v^2| = \log|x| + C'$

Rearrange the terms:

$-\log|1 - v^2| - \log|x| = C'$

$-(\log|1 - v^2| + \log|x|) = C'$

$- \log(|x(1 - v^2)|) = C'$

$\log(|x(1 - v^2)|) = -C'$

Exponentiate both sides with base $e$:

$|x(1 - v^2)| = e^{-C'}$

$x(1 - v^2) = A$

Substitute back $v = \frac{y}{x}$ into equation (iii):

$x \left(1 - \left(\frac{y}{x}\right)^2\right) = A$

$x \left(1 - \frac{y^2}{x^2}\right) = A$

$x \left(\frac{x^2 - y^2}{x^2}\right) = A$

$\frac{x^2 - y^2}{x} = A$

This is the general solution of the differential equation for $x \neq 0$ and $y \neq \pm x$.

Now, we use the condition that the curve passes through the point $(2, 1)$ to find the value of the constant $A$. Substitute $x=2$ and $y=1$ into equation (iv):

$2^2 - 1^2 = A(2)$

$4 - 1 = 2A$

$3 = 2A$

Substitute the value of $A = \frac{3}{2}$ back into the general solution (iv) to obtain the equation of the curve:

$x^2 - y^2 = \frac{3}{2} x$

Multiply by 2 to remove the fraction:

$2(x^2 - y^2) = 3x$

$2x^2 - 2y^2 = 3x$

Final Answer:

The equation of the curve passing through $(2, 1)$ is $x^2 - y^2 = \frac{3}{2} x$ or $2x^2 - 2y^2 = 3x$.

Given:

The slope of the tangent to the curve at any point $(x, y)$ is $\frac{y − 1}{x^2 + x}$.

The curve passes through the point $(1, 0)$.

To Find:

The equation of the curve.

Solution:

The slope of the tangent to the curve at any point $(x, y)$ is given by $\frac{dy}{dx}$. So, the differential equation of the curve is:

This is a variable separable differential equation. We can separate the terms involving $y$ and $dy$ from the terms involving $x$ and $dx$. Assuming $y-1 \neq 0$ and $x^2+x \neq 0$:

$\frac{dy}{y − 1} = \frac{dx}{x^2 + x}$

$\frac{dy}{y − 1} = \frac{dx}{x(x + 1)}$

Integrate both sides of the separated equation:

$\int \frac{dy}{y − 1} = \int \frac{dx}{x(x + 1)}$

For the integral on the left-hand side, let $u = y-1$, then $du = dy$. $\int \frac{du}{u} = \log|u| = \log|y-1|$.

For the integral on the right-hand side, we use partial fraction decomposition:

$\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}$

$1 = A(x+1) + Bx$

Setting $x=0$, we get $1 = A(1) + B(0) \implies A = 1$.

Setting $x=-1$, we get $1 = A(0) + B(-1) \implies B = -1$.

So, $\frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1}$.

The integral becomes:

$\int \left(\frac{1}{x} - \frac{1}{x+1}\right) dx = \int \frac{1}{x} dx - \int \frac{1}{x+1} dx = \log|x| - \log|x+1| + C'$

$= \log\left|\frac{x}{x+1}\right| + C'$

Equating the results of the integrals:

$\log|y-1| = \log\left|\frac{x}{x+1}\right| + C$

Now, we use the condition that the curve passes through the point $(1, 0)$. Substitute $x=1$ and $y=0$ into equation (ii):

$\log|0-1| = \log\left|\frac{1}{1+1}\right| + C$

$\log|-1| = \log\left|\frac{1}{2}\right| + C$

$\log(1) = \log\left(\frac{1}{2}\right) + C$

Since $\log(1) = 0$:

$0 = \log\left(\frac{1}{2}\right) + C$

$C = -\log\left(\frac{1}{2}\right)$

$C = -(\log 1 - \log 2)$

$C = -(0 - \log 2)$

Substitute the value of $C = \log 2$ back into the general solution (ii) to obtain the equation of the curve:

$\log|y-1| = \log\left|\frac{x}{x+1}\right| + \log 2$

Using the property $\log a + \log b = \log (ab)$:

$\log|y-1| = \log\left| \frac{x}{x+1} \cdot 2 \right|$

$\log|y-1| = \log\left|\frac{2x}{x+1}\right|$

Exponentiate both sides with base $e$ to remove the logarithm:

$|y-1| = \left|\frac{2x}{x+1}\right|$

Since the curve passes through $(1, 0)$, where $y-1 = 0-1 = -1$ (negative) and $\frac{2x}{x+1} = \frac{2(1)}{1+1} = \frac{2}{2} = 1$ (positive), the absolute values must be handled carefully to ensure consistency. For the specific branch containing $(1,0)$, we have $y-1$ negative and $\frac{2x}{x+1}$ positive (for $x>0$).

So, we can write:

$-(y-1) = \frac{2x}{x+1}$

$1 - y = \frac{2x}{x+1}$

Solve for $y$:

$y = 1 - \frac{2x}{x+1}$

$y = \frac{x+1}{x+1} - \frac{2x}{x+1}$

$y = \frac{x+1-2x}{x+1}$

Final Answer:

The equation of the curve through the point $(1, 0)$ is $y = \frac{1-x}{x+1}$ or $(y-1)(x+1) = 2x$ (from $(y-1) = -\frac{2x}{x+1}$).

Given:

The curve passes through the origin $(0, 0)$.

The slope of the tangent to the curve at any point $(x, y)$ is equal to the square of the difference of the abscissa and ordinate of the point.

To Find:

The equation of the curve.

Solution:

The slope of the tangent to the curve at any point $(x, y)$ is given by $\frac{dy}{dx}$. According to the problem statement, this slope is equal to the square of the difference of the abscissa ($x$) and the ordinate ($y$).

So, the differential equation of the curve is:

This is a differential equation that can be solved using a substitution.

Let $z = x - y$.

Differentiate both sides with respect to $x$:

$\frac{dz}{dx} = \frac{d}{dx}(x - y)$

$\frac{dz}{dx} = 1 - \frac{dy}{dx}$

From this, we can express $\frac{dy}{dx}$ as:

Substitute equation (ii) and $z = x-y$ into the original differential equation (i):

$1 - \frac{dz}{dx} = z^2$

Rearrange the equation to separate variables:

$1 - z^2 = \frac{dz}{dx}$

$dx = \frac{dz}{1 - z^2}$

Integrate both sides of the separated equation:

$\int dx = \int \frac{dz}{1 - z^2}$

The integral on the left side is $\int dx = x + C'$.

The integral on the right side is a standard integral: $\int \frac{dz}{1 - z^2} = \frac{1}{2} \log\left|\frac{1+z}{1-z}\right|$.

Equating the results of the integrals and adding the arbitrary constant of integration $C$:

Substitute back $z = x-y$ into equation (iii):

Now, we use the initial condition that the curve passes through the origin $(0, 0)$. Substitute $x=0$ and $y=0$ into the general solution (iv):

$0 = \frac{1}{2} \log\left|\frac{1+0-0}{1-0+0}\right| + C$

$0 = \frac{1}{2} \log\left|\frac{1}{1}\right| + C$

$0 = \frac{1}{2} \log(1) + C$

Since $\log(1) = 0$:

$0 = \frac{1}{2} (0) + C$

Substitute the value of $C = 0$ back into the general solution (iv) to obtain the particular solution, which is the equation of the curve passing through the origin:

$x = \frac{1}{2} \log\left|\frac{1+x-y}{1-x+y}\right|$

Multiply both sides by 2:

$2x = \log\left|\frac{1+x-y}{1-x+y}\right|$

Exponentiate both sides with base $e$:

$e^{2x} = \left|\frac{1+x-y}{1-x+y}\right|$

Since the curve passes through the origin $(0,0)$, we can check the sign of the term inside the absolute value at $(0,0)$: $\frac{1+0-0}{1-0+0} = 1$, which is positive. Assuming the curve does not cross the lines where the numerator or denominator is zero (i.e., $1+x-y=0$ or $1-x+y=0$), the sign remains positive for the branch of the curve passing through the origin. Thus, we can remove the absolute value:

Final Answer:

The equation of the curve passing through the origin is $e^{2x} = \frac{1+x-y}{1-x+y}$.

Given:

A curve passes through the point $(1, 1)$.

The tangent at any point $P(x, y)$ on the curve meets the coordinate axes at A and B such that P is the mid-point of AB.

To Find:

The equation of the curve.

Solution:

Let $P(x, y)$ be a point on the curve. Let the tangent to the curve at P meet the x-axis at point $A(a, 0)$ and the y-axis at point $B(0, b)$.

According to the problem, P is the mid-point of the line segment AB.

Using the mid-point formula, the coordinates of P are the average of the coordinates of A and B:

$(x, y) = \left(\frac{a+0}{2}, \frac{0+b}{2}\right)$

Equating the coordinates:

From equations (i) and (ii), we find the coordinates of A and B in terms of $x$ and $y$:

$a = 2x$

$b = 2y$

So, the points are $A(2x, 0)$ and $B(0, 2y)$.

The tangent line passes through the points $A(2x, 0)$ and $B(0, 2y)$. The slope of the tangent line at $P(x, y)$ is given by $\frac{dy}{dx}$. We can calculate this slope using points A and B:

Slope $= \frac{y_2 - y_1}{x_2 - x_1} = \frac{2y - 0}{0 - 2x} = \frac{2y}{-2x} = -\frac{y}{x}$

Thus, the differential equation of the curve is:

This is a variable separable differential equation. Assuming $y \neq 0$ and $x \neq 0$, we can separate the variables:

$\frac{dy}{y} = -\frac{dx}{x}$

Integrate both sides:

$\int \frac{dy}{y} = \int -\frac{dx}{x}$

$\log|y| = -\log|x| + C'$

Rearrange the equation:

$\log|y| + \log|x| = C'$

$\log|xy| = C'$

Exponentiate both sides with base $e$:

$|xy| = e^{C'}$

$xy = \pm e^{C'}$

Let $K = \pm e^{C'}$, where $K$ is an arbitrary non-zero constant.

We are given that the curve passes through the point $(1, 1)$. Substitute these coordinates into the general solution to find the value of $K$:

$(1)(1) = K$

Substitute the value of $K=1$ back into the general solution to obtain the particular solution:

Note: The case $x=0$ or $y=0$ was excluded during separation. However, the solution $xy=1$ does not contain any points where $x=0$ or $y=0$.

Final Answer:

The equation of the curve passing through the point (1, 1) is $xy = 1$.

Given:

The differential equation $x \frac{dy}{dx} = y (\log y – \log x + 1)$.

To Solve:

Find the general solution of the given differential equation.

Solution:

The given differential equation is:

Assuming $x \neq 0$, we can divide both sides by $x$:

$\frac{dy}{dx} = \frac{y}{x} (\log y – \log x + 1)$

Using the property of logarithms $\log y – \log x = \log\left(\frac{y}{x}\right)$ (assuming $x>0$ and $y>0$), we get:

This is a homogeneous differential equation because the right-hand side is a function of $\frac{y}{x}$.

We use the substitution $y = vx$, where $v$ is a function of $x$.

Differentiate $y = vx$ with respect to $x$ using the product rule:

Substitute equation (iii) and $y = vx$ (so $\frac{y}{x} = v$) into equation (ii):

$v + x \frac{dv}{dx} = v (\log v + 1)$

$v + x \frac{dv}{dx} = v \log v + v$

Subtract $v$ from both sides:

This is a variable separable differential equation. Assuming $x \neq 0$ and $v \log v \neq 0$ (i.e., $v \neq 1$ and $v > 0$), we can separate the variables $v$ and $x$:

$\frac{dv}{v \log v} = \frac{dx}{x}$

Integrate both sides of the separated equation:

$\int \frac{dv}{v \log v} = \int \frac{dx}{x}$

Evaluate the integral on the left-hand side, $\int \frac{dv}{v \log v}$. Let $u = \log v$. Then $du = \frac{1}{v} dv$. The integral becomes:

$\int \frac{du}{u} = \log|u|$

Substitute back $u = \log v$:

$\int \frac{dv}{v \log v} = \log|\log v|$

Evaluate the integral on the right-hand side:

$\int \frac{dx}{x} = \log|x| + C'$

Equating the results of the integrals:

$\log|\log v| = \log|x| + C$

Rearrange equation (vi):

$\log|\log v| - \log|x| = C$

$\log\left|\frac{\log v}{x}\right| = C$

Exponentiate both sides with base $e$:

$\left|\frac{\log v}{x}\right| = e^C$

$\frac{\log v}{x} = \pm e^C$

Let $K = \pm e^C$, where $K$ is a non-zero arbitrary constant.

$\log v = Kx$

Exponentiate both sides with base $e$ again:

$v = e^{Kx}$

Substitute back $v = \frac{y}{x}$:

$\frac{y}{x} = e^{Kx}$

$y = x e^{Kx}$

Note: We assumed $v \log v \neq 0$. If $v \log v = 0$, then either $v=0$ or $v=1$. If $v=y/x=0$, then $y=0$. Substituting $y=0$ into the original DE gives $x \frac{d(0)}{dx} = 0 (\log 0 - \log x + 1)$, which is $0=0$. So $y=0$ is a solution. If $v=y/x=1$, then $y=x$. Substituting $y=x$ gives $x \frac{d(x)}{dx} = x(\log x - \log x + 1)$, which is $x=x$. So $y=x$ is also a solution. The general solution $y = x e^{Kx}$ includes $y=0$ if $K$ can be negative infinity in some limiting sense, and includes $y=x$ if $K=0$. However, $K$ was defined as non-zero. The standard method provides $y = x e^{Kx}$ for $K \neq 0$ and $y=x$ and $y=0$ as potential singular solutions.

Final Answer:

The general solution of the differential equation is $y = x e^{Kx}$, where $\mathbf{K}$ is an arbitrary constant.

The given differential equation is:

$\left( \frac{d^2y}{dx^2} \right)^2 + \left( \frac{dy}{dx} \right)^2 = x \sin \left( \frac{dy}{dx} \right)$

To find the degree of a differential equation, we first need to ensure it is a polynomial equation in its derivatives. The degree is then the highest power of the highest order derivative present in the equation.

In the given equation, the highest order derivative is $\frac{d^2y}{dx^2}$. The order of the differential equation is 2.

However, the term $\sin \left( \frac{dy}{dx} \right)$ involves a transcendental function of a derivative. For the degree of a differential equation to be defined, the equation must be expressible as a polynomial in the derivatives $\frac{dy}{dx}, \frac{d^2y}{dx^2}, \dots, \frac{d^ny}{dx^n}$. Since the given equation involves $\sin \left( \frac{dy}{dx} \right)$, which cannot be expressed as a finite polynomial in $\frac{dy}{dx}$, the equation is not a polynomial in its derivatives.

Therefore, the degree of the differential equation is not defined.

The correct option is (D) not defined.

The given differential equation is:

$\left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^{\frac{3}{2}} = \frac{d^2y}{dx^2}$

To find the degree of a differential equation, it must be a polynomial equation in its derivatives. The given equation has a fractional power ($\frac{3}{2}$). To make it a polynomial in derivatives, we need to remove the fractional power by squaring both sides of the equation.

Squaring both sides, we get:

$\left( \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^{\frac{3}{2}} \right)^2 = \left( \frac{d^2y}{dx^2} \right)^2$

$\left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^{3} = \left( \frac{d^2y}{dx^2} \right)^2$

Now, the equation is a polynomial in terms of the derivatives $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$.

The order of the differential equation is the order of the highest derivative present, which is $\frac{d^2y}{dx^2}$. So, the order is 2.

The degree of the differential equation is the highest power of the highest order derivative present in the polynomial form of the equation. In the equation $\left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^{3} = \left( \frac{d^2y}{dx^2} \right)^2$, the highest order derivative is $\frac{d^2y}{dx^2}$, and its power is 2.

Therefore, the degree of the differential equation is 2.

The correct option is (D) 2.

The given differential equation is:

$\frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^{\frac{1}{4}} + x^{\frac{1}{5}} = 0$

The order of a differential equation is the order of the highest derivative present in the equation.

In the given equation, the highest derivative is $\frac{d^2y}{dx^2}$, which is the second derivative. Thus, the order of the differential equation is 2.

The degree of a differential equation is the highest power of the highest order derivative present in the differential equation when it has been made free from radicals and fractions, as far as the derivatives are concerned. The degree is defined only if the differential equation can be expressed as a polynomial in its derivatives.

The given equation involves a fractional power $\frac{1}{4}$ on the derivative $\frac{dy}{dx}$. To clear this fractional power, we can isolate the term $\left( \frac{dy}{dx} \right)^{\frac{1}{4}}$:

$\left( \frac{dy}{dx} \right)^{\frac{1}{4}} = - \frac{d^2y}{dx^2} - x^{\frac{1}{5}}$

Now, raise both sides of the equation to the power of 4:

$\left( \left( \frac{dy}{dx} \right)^{\frac{1}{4}} \right)^4 = \left( - \frac{d^2y}{dx^2} - x^{\frac{1}{5}} \right)^4$

$\frac{dy}{dx} = \left( \frac{d^2y}{dx^2} + x^{\frac{1}{5}} \right)^4$

Expanding the right side using the binomial theorem, we get terms involving powers of $\frac{d^2y}{dx^2}$ and $x^{\frac{1}{5}}$:

$\frac{dy}{dx} = \left( \frac{d^2y}{dx^2} \right)^4 + 4 \left( \frac{d^2y}{dx^2} \right)^3 x^{\frac{1}{5}} + 6 \left( \frac{d^2y}{dx^2} \right)^2 x^{\frac{2}{5}} + 4 \left( \frac{d^2y}{dx^2} \right)^1 x^{\frac{3}{5}} + x^{\frac{4}{5}}$

This equation is a polynomial in terms of the derivatives $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$. The coefficients of the polynomial terms involving derivatives are functions of $x$, specifically $1$, $4x^{\frac{1}{5}}$, $6x^{\frac{2}{5}}$, $4x^{\frac{3}{5}}$.

According to the standard definition, the degree is the highest power of the highest order derivative in this polynomial form. The highest order derivative is $\frac{d^2y}{dx^2}$, and its highest power in the expanded equation is 4.

Thus, based on the standard definition, the order is 2 and the degree is 4.

However, observing the given options, a degree of 4 is not listed. Option (A) suggests "not defined" for the degree.

The degree is "not defined" if the differential equation cannot be expressed as a polynomial in its derivatives (e.g., if a derivative is inside a transcendental function like $\sin(y')$ or $e^{y''}$). We have shown that the equation *can* be written as a polynomial in derivatives.

A possible reason for the option "not defined" (though not standard) could be the presence of non-polynomial terms in $x$ (like $x^{\frac{1}{5}}, x^{\frac{2}{5}}$ etc.) as coefficients after clearing the fractional powers of the derivatives. If the definition of degree requires the coefficients to be, say, polynomials in $x$ or constants, then the degree might be considered "not defined" in this specific context.

Given the options, and assuming the question intends one of the provided choices, the most likely intended answer is that the degree is not defined, despite the equation being expressible as a polynomial in derivatives with coefficients that are functions of $x$.

Therefore, the order is 2 and the degree is not defined.

The correct option is (A) 2 and not defined.

The given function is:

$y = e^{-x}(A \cos x + B \sin x)$

We need to find the differential equation satisfied by this function. We will calculate the first and second derivatives of $y$ with respect to $x$.

First derivative $\frac{dy}{dx}$:

Using the product rule, $\frac{d}{dx}(uv) = u'v + uv'$, where $u = e^{-x}$ and $v = A \cos x + B \sin x$.

$\frac{du}{dx} = \frac{d}{dx}(e^{-x}) = -e^{-x}$

$\frac{dv}{dx} = \frac{d}{dx}(A \cos x + B \sin x) = -A \sin x + B \cos x$

So,

$\frac{dy}{dx} = (-e^{-x})(A \cos x + B \sin x) + e^{-x}(-A \sin x + B \cos x)$

$\frac{dy}{dx} = -e^{-x}(A \cos x + B \sin x) + e^{-x}(B \cos x - A \sin x)$

We can notice that $-e^{-x}(A \cos x + B \sin x) = -y$. So,

Second derivative $\frac{d^2y}{dx^2}$:

Differentiate equation (i) with respect to $x$:

$\frac{d^2y}{dx^2} = \frac{d}{dx}(-y) + \frac{d}{dx}(e^{-x}(B \cos x - A \sin x))$

$\frac{d^2y}{dx^2} = -\frac{dy}{dx} + \frac{d}{dx}(e^{-x})(B \cos x - A \sin x) + e^{-x}\frac{d}{dx}(B \cos x - A \sin x)$

$\frac{d^2y}{dx^2} = -\frac{dy}{dx} + (-e^{-x})(B \cos x - A \sin x) + e^{-x}(-B \sin x - A \cos x)$

$\frac{d^2y}{dx^2} = -\frac{dy}{dx} - e^{-x}(B \cos x - A \sin x) - e^{-x}(A \cos x + B \sin x)$

From equation (i), we have $e^{-x}(B \cos x - A \sin x) = \frac{dy}{dx} + y$.

Also, $e^{-x}(A \cos x + B \sin x) = y$.

Substitute these back into the expression for $\frac{d^2y}{dx^2}$:

$\frac{d^2y}{dx^2} = -\frac{dy}{dx} - (\frac{dy}{dx} + y) - y$

$\frac{d^2y}{dx^2} = -\frac{dy}{dx} - \frac{dy}{dx} - y - y$

$\frac{d^2y}{dx^2} = -2 \frac{dy}{dx} - 2y$

Rearranging the terms, we get:

$\frac{d^2y}{dx^2} + 2 \frac{dy}{dx} + 2y = 0$

This is a linear homogeneous differential equation with constant coefficients.

Comparing this with the given options, we find that it matches option (C).

The correct option is (C) $\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 2y = 0$.

Given:

$y = A \cos αx + B \sin αx$, where A and B are arbitrary constants.

To Find:

The differential equation satisfied by the given function.

Solution:

We are given the function:

$y = A \cos αx + B \sin αx$

We need to find the first and second derivatives of $y$ with respect to $x$ to eliminate the two arbitrary constants A and B.

Differentiate $y$ with respect to $x$ to find $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{d}{dx}(A \cos αx + B \sin αx)$

$\frac{dy}{dx} = A \frac{d}{dx}(\cos αx) + B \frac{d}{dx}(\sin αx)$

Using the chain rule, $\frac{d}{dx}(\cos αx) = -\sin αx \cdot α$ and $\frac{d}{dx}(\sin αx) = \cos αx \cdot α$.

So,

$\frac{dy}{dx} = A (-\sin αx \cdot α) + B (\cos αx \cdot α)$

$\frac{dy}{dx} = -Aα \sin αx + Bα \cos αx$

Differentiate $\frac{dy}{dx}$ with respect to $x$ to find $\frac{d^2y}{dx^2}$:

$\frac{d^2y}{dx^2} = \frac{d}{dx}(-Aα \sin αx + Bα \cos αx)$

$\frac{d^2y}{dx^2} = -Aα \frac{d}{dx}(\sin αx) + Bα \frac{d}{dx}(\cos αx)$

Using the chain rule again,

$\frac{d^2y}{dx^2} = -Aα (\cos αx \cdot α) + Bα (-\sin αx \cdot α)$

$\frac{d^2y}{dx^2} = -Aα^2 \cos αx - Bα^2 \sin αx$

Factor out $-α^2$ from the expression for $\frac{d^2y}{dx^2}$:

$\frac{d^2y}{dx^2} = -α^2 (A \cos αx + B \sin αx)$

Notice that the term in the parenthesis is the original function $y$. Substitute $y$ back into the equation:

$\frac{d^2y}{dx^2} = -α^2 y$

Rearrange the equation to get the differential equation:

$\frac{d^2y}{dx^2} + α^2 y = 0$

This is the required differential equation satisfied by the given function.

Comparing this equation with the given options:

(A) $\frac{d^2y}{dx^2} − α^2y = 0$

(B) $\frac{d^2y}{dx^2} + α^2y = 0$

(C) $\frac{d^2y}{dx^2} + αy = 0$

(D) $\frac{d^2y}{dx^2} − αy = 0$

The derived equation matches option (B).

The correct option is (B) $\frac{d^2y}{dx^2} + α^2y = 0$.

The given differential equation is:

$x \, dy - y \, dx = 0$

We can rewrite the equation as:

$x \, dy = y \, dx$

Assuming $x \neq 0$ and $y \neq 0$, we can separate the variables:

$\frac{dy}{y} = \frac{dx}{x}$

Now, integrate both sides of the equation:

$\int \frac{1}{y} \, dy = \int \frac{1}{x} \, dx$

$\ln|y| = \ln|x| + C_1$

where $C_1$ is the constant of integration.

We can rewrite the constant $C_1$ as $\ln|C|$, where $C$ is a non-zero constant.

$\ln|y| = \ln|x| + \ln|C|$

$\ln|y| = \ln|Cx|$

Exponentiating both sides:

$|y| = |Cx|$

$y = \pm Cx$

Let $K = \pm C$. Since $C$ is a non-zero arbitrary constant, $K$ is also a non-zero arbitrary constant. So, the general solution is:

$y = Kx$, where $K \neq 0$.

We should also check the cases where $x=0$ or $y=0$. If $y=0$, then $dy=0$, and the original equation becomes $x(0) - 0(dx) = 0$, which is $0=0$. Thus, $y=0$ is a solution, and it is included in $y=Kx$ when $K=0$. If $x=0$, then $dx=0$, and the original equation becomes $0(dy) - y(0) = 0$, which is $0=0$. Thus, $x=0$ is also a solution. The family $y=Kx$ represents all lines passing through the origin, including the x-axis ($K=0$), and as $K \to \pm \infty$, it approaches the y-axis. Geometrically, the solutions are all straight lines passing through the origin $(0,0)$.

The equation $y = Kx$ represents a family of straight lines that pass through the origin $(0,0)$.

Comparing this geometric description with the given options:

(A) a rectangular hyperbola: Equations like $xy = k$ or $x^2 - y^2 = k$.

(B) parabola whose vertex is at origin: Equations like $y = ax^2$ or $x = ay^2$.

(C) straight line passing through origin: Equations like $y = mx$. This matches $y=Kx$.

(D) a circle whose centre is at origin: Equation like $x^2 + y^2 = r^2$.

The solution represents a straight line passing through the origin.

The correct option is (C) straight line passing through origin.

Given:

The differential equation $\cos x \frac{dy}{dx} + y \sin x = 1$.

To Find:

The integrating factor of the differential equation.

Solution:

The given differential equation is:

$\cos x \frac{dy}{dx} + y \sin x = 1$

This is a first-order linear differential equation. The standard form of a first-order linear differential equation is:

$\frac{dy}{dx} + P(x)y = Q(x)$

To convert the given equation into the standard form, divide the entire equation by $\cos x$, assuming $\cos x \neq 0$:

$\frac{\cos x}{\cos x} \frac{dy}{dx} + \frac{\sin x}{\cos x} y = \frac{1}{\cos x}$

$\frac{dy}{dx} + \tan x \cdot y = \sec x$

Now, comparing this equation with the standard form, we can identify $P(x)$ and $Q(x)$:

$P(x) = \tan x$

$Q(x) = \sec x$

The integrating factor (IF) for a first-order linear differential equation is given by the formula:

$IF = e^{\int P(x) dx}$

Substitute $P(x) = \tan x$ into the formula:

$IF = e^{\int \tan x \, dx}$

Now, evaluate the integral $\int \tan x \, dx$:

$\int \tan x \, dx = \int \frac{\sin x}{\cos x} \, dx$

Let $u = \cos x$, then $du = -\sin x \, dx$. So, $\sin x \, dx = -du$.

$\int \frac{\sin x}{\cos x} \, dx = \int \frac{-du}{u} = -\int \frac{1}{u} \, du = -\ln|u| + C'$

Substitute back $u = \cos x$:

$\int \tan x \, dx = -\ln|\cos x| + C'$

Using the property of logarithms, $-\ln|\cos x| = \ln(|\cos x|^{-1}) = \ln|\frac{1}{\cos x}| = \ln|\sec x|$.

For the integrating factor, we only need one particular antiderivative, so we can take the constant of integration $C'$ to be 0.

$\int \tan x \, dx = \ln|\sec x|$

Substitute this result back into the formula for the integrating factor:

$IF = e^{\ln|\sec x|}$

Using the property $e^{\ln a} = a$ for $a>0$. We can assume $\sec x > 0$ in the interval where the solution is sought. Even if $\sec x < 0$, the absolute value makes it positive, and $e^{\ln|\sec x|} = |\sec x|$. However, the standard integrating factor is often taken without the absolute value sign, as multiplying by it works even if it's negative. For options provided, $\sec x$ is a direct match.

$IF = \sec x$

The integrating factor of the given differential equation is $\sec x$.

Comparing this with the given options:

(A) cos x

(B) tan x

(C) sec x

(D) sin x

The calculated integrating factor matches option (C).

The correct option is (C) sec x.

Given:

The differential equation $\tan y \sec^2 x \, dx + \tan x \sec^2 y \, dy = 0$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is:

$\tan y \sec^2 x \, dx + \tan x \sec^2 y \, dy = 0$

This is a variable separable differential equation. To separate the variables, divide the entire equation by $(\tan x \tan y)$ (assuming $\tan x \neq 0$ and $\tan y \neq 0$).

$\frac{\tan y \sec^2 x}{\tan x \tan y} \, dx + \frac{\tan x \sec^2 y}{\tan x \tan y} \, dy = 0$

$\frac{\sec^2 x}{\tan x} \, dx + \frac{\sec^2 y}{\tan y} \, dy = 0$

Now, integrate both sides of the separated equation:

$\int \frac{\sec^2 x}{\tan x} \, dx + \int \frac{\sec^2 y}{\tan y} \, dy = \int 0$

Let's evaluate the first integral $\int \frac{\sec^2 x}{\tan x} \, dx$.

Let $u = \tan x$. Then $du = \sec^2 x \, dx$.

The integral becomes $\int \frac{1}{u} \, du = \ln|u|$. Substituting back $u = \tan x$, we get $\ln|\tan x|$.

Let's evaluate the second integral $\int \frac{\sec^2 y}{\tan y} \, dy$.

Let $v = \tan y$. Then $dv = \sec^2 y \, dy$.

The integral becomes $\int \frac{1}{v} \, dv = \ln|v|$. Substituting back $v = \tan y$, we get $\ln|\tan y|$.

The integral of the right side is a constant. Let the constant of integration be $\ln|k|$, where $k$ is an arbitrary constant (assuming $k \neq 0$).

So, the integrated equation is:

$\ln|\tan x| + \ln|\tan y| = \ln|k|$

Using the property of logarithms, $\ln a + \ln b = \ln (ab)$:

$\ln|\tan x \cdot \tan y| = \ln|k|$

Exponentiating both sides:

$|\tan x \cdot \tan y| = |k|$

$\tan x \cdot \tan y = \pm k$

Let $C = \pm k$. Since $k$ is an arbitrary non-zero constant, $C$ is also a non-zero arbitrary constant. If we consider the cases where $\tan x = 0$ or $\tan y = 0$, these correspond to $x = n\pi$ or $y = m\pi$ for integers $n, m$. In these cases, the original equation is satisfied (0 = 0). The solution $\tan x \cdot \tan y = 0$ is included in $\tan x \cdot \tan y = C$ when $C=0$. Thus, the solution can be written as:

$\tan x \cdot \tan y = C$, where $C$ is an arbitrary constant.

Comparing this solution with the given options, where the constant is denoted by $k$:

(A) $\tan x + \tan y = k$

(B) $\tan x – \tan y = k$

(C) $\frac{\tan x}{\tan y} = k$ (equivalent to $\tan x = k \tan y$)

(D) $\tan x \cdot \tan y = k$

Our derived solution matches option (D).

The correct option is (D) tan x . tan y = k.

Given:

The family of curves is given by the equation:

$y = Ax + A^3$

where A is an arbitrary constant.

To Find:

The degree of the differential equation representing this family of curves.

Solution:

The given equation contains one arbitrary constant A. Therefore, the order of the differential equation will be 1.

To obtain the differential equation, we need to eliminate the arbitrary constant A by differentiating the given equation with respect to $x$.

Differentiating $y = Ax + A^3$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(Ax + A^3)$

$\frac{dy}{dx} = A \frac{d}{dx}(x) + \frac{d}{dx}(A^3)$

Since A is a constant, $A^3$ is also a constant, and its derivative is 0. $\frac{d}{dx}(x) = 1$.

So, $\frac{dy}{dx} = A(1) + 0$

$\frac{dy}{dx} = A$

Now, substitute the value of A from this derivative back into the original equation $y = Ax + A^3$.

Substitute $A = \frac{dy}{dx}$ into $y = Ax + A^3$:

$y = \left(\frac{dy}{dx}\right) x + \left(\frac{dy}{dx}\right)^3$

This is the differential equation representing the given family of curves.

The differential equation is $y = x \frac{dy}{dx} + \left(\frac{dy}{dx}\right)^3$.

To find the degree, we look at the highest power of the highest order derivative in the polynomial form of the differential equation. The highest order derivative is $\frac{dy}{dx}$, which is a first-order derivative. The powers of $\frac{dy}{dx}$ in the equation are 1 (in $x \frac{dy}{dx}$) and 3 (in $\left(\frac{dy}{dx}\right)^3$).

The highest power of the highest order derivative ($\frac{dy}{dx}$) is 3.

Therefore, the degree of the differential equation is 3.

Comparing this with the given options:

(A) 1

(B) 2

(C) 3

(D) 4

The calculated degree matches option (C).

The correct option is (C) 3.

Given:

The differential equation $x \frac{dy}{dx} - y = x^4 – 3x$.

To Find:

The integrating factor of the differential equation.

Solution:

The given differential equation is:

$x \frac{dy}{dx} - y = x^4 – 3x$

This is a first-order linear differential equation. The standard form of a first-order linear differential equation is:

$\frac{dy}{dx} + P(x)y = Q(x)$

To convert the given equation into the standard form, divide the entire equation by the coefficient of $\frac{dy}{dx}$, which is $x$ (assuming $x \neq 0$):

$\frac{x}{x} \frac{dy}{dx} - \frac{y}{x} = \frac{x^4 – 3x}{x}$

$\frac{dy}{dx} - \frac{1}{x} y = x^3 – 3$

Now, comparing this equation with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we can identify $P(x)$ and $Q(x)$:

$P(x) = -\frac{1}{x}$

$Q(x) = x^3 – 3$

The integrating factor (IF) for a first-order linear differential equation is given by the formula:

$IF = e^{\int P(x) dx}$

Substitute $P(x) = -\frac{1}{x}$ into the formula:

$IF = e^{\int -\frac{1}{x} \, dx}$

Now, evaluate the integral $\int -\frac{1}{x} \, dx$:

$\int -\frac{1}{x} \, dx = - \int \frac{1}{x} \, dx = -\ln|x| + C'$

Using the property of logarithms, $-\ln|x| = \ln(|x|^{-1}) = \ln|\frac{1}{x}|$.

For the integrating factor, we only need one particular antiderivative, so we can take the constant of integration $C'$ to be 0.

$\int -\frac{1}{x} \, dx = \ln|\frac{1}{x}|$

Substitute this result back into the formula for the integrating factor:

$IF = e^{\ln|\frac{1}{x}|}$

Using the property $e^{\ln a} = a$ for $a>0$. Assuming $x \neq 0$, $\frac{1}{x} \neq 0$. We can take $IF = \frac{1}{x}$ (valid for $x>0$ or $x<0$).

$IF = \frac{1}{x}$

The integrating factor of the given differential equation is $\frac{1}{x}$.

Comparing this with the given options:

(A) x

(B) log x

(C) $\frac{1}{x}$

(D) – x

The calculated integrating factor matches option (C).

The correct option is (C) $\frac{1}{x}$.

Given:

The differential equation $\frac{dy}{dx} - y = 1$ with the initial condition $y(0) = 1$.

To Find:

The particular solution of the differential equation satisfying the initial condition.

Solution:

The given differential equation is:

$\frac{dy}{dx} - y = 1$

This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.

Comparing the given equation with the standard form, we have:

$P(x) = -1$

$Q(x) = 1$

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

$IF = e^{\int (-1) \, dx}$

$IF = e^{-x}$

The general solution of a first-order linear differential equation is given by:

$y \cdot (IF) = \int Q(x) \cdot (IF) \, dx + C$

Substitute the values of $y$, $IF$, and $Q(x)$:

$y \cdot e^{-x} = \int 1 \cdot e^{-x} \, dx + C$

$y e^{-x} = \int e^{-x} \, dx + C$

Evaluate the integral $\int e^{-x} \, dx$:

$\int e^{-x} \, dx = -e^{-x} + C'$

So,

$y e^{-x} = -e^{-x} + C$

To find the general solution explicitly for $y$, multiply both sides by $e^x$:

$y = \frac{-e^{-x} + C}{e^{-x}}$

$y = -1 + C e^x$

$y = C e^x - 1$

Now, we use the initial condition $y(0) = 1$ to find the value of the arbitrary constant C.

Substitute $x=0$ and $y=1$ into the general solution:

$1 = C e^0 - 1$

$1 = C(1) - 1$

$1 = C - 1$

$C = 1 + 1$

$C = 2$

Substitute the value of C back into the general solution to obtain the particular solution:

$y = 2 e^x - 1$

Comparing this particular solution with the given options:

(A) $xy = – e^{x}$ (equivalent to $y = -\frac{e^x}{x}$)

(B) $xy = – e^{–x}$ (equivalent to $y = -\frac{e^{-x}}{x}$)

(C) $xy = – 1$ (equivalent to $y = -\frac{1}{x}$)

(D) $y = 2 e^{x} – 1$

Our derived particular solution matches option (D).

The correct option is (D) y = 2 e^x – 1.

Given:

The differential equation $\frac{dy}{dx} = \frac{y + 1}{x − 1}$ and the initial condition $y(1) = 2$.

To Find:

The number of solutions of the differential equation satisfying the initial condition.

Solution:

The given differential equation is:

$\frac{dy}{dx} = \frac{y + 1}{x − 1}$

This is a first-order differential equation. We can check the existence and uniqueness of a solution passing through a given point $(x_0, y_0)$ using the Existence and Uniqueness Theorem for first-order differential equations of the form $\frac{dy}{dx} = f(x, y)$.

In this case, $f(x, y) = \frac{y + 1}{x − 1}$. The initial condition is given at $(x_0, y_0) = (1, 2)$.

The function $f(x, y) = \frac{y+1}{x-1}$ is defined for all $(x, y)$ where $x \neq 1$.

The initial condition is given at the point $(1, 2)$. At this point, $x = 1$.

Substitute the point $(1, 2)$ into the function $f(x, y)$:

$f(1, 2) = \frac{2 + 1}{1 - 1} = \frac{3}{0}$

Since $f(1, 2)$ is undefined, the function $f(x, y)$ is not continuous at the point $(1, 2)$.

The standard Existence and Uniqueness Theorem states that if $f(x, y)$ and $\frac{\partial f}{\partial y}$ are continuous in a rectangular region containing $(x_0, y_0)$, then there exists a unique solution through $(x_0, y_0)$. Since $f(x, y)$ is not even defined at $(1, 2)$, the conditions of this theorem are not met.

Let's try to solve the differential equation by separating variables to see if we encounter any issue:

$\frac{dy}{y+1} = \frac{dx}{x-1}$

Integrate both sides:

$\int \frac{1}{y+1} \, dy = \int \frac{1}{x-1} \, dx$

$\log|y+1| = \log|x-1| + C_1$

where $C_1$ is the constant of integration.

Let $C_1 = \log|C|$, where $C$ is a non-zero arbitrary constant.

$\log|y+1| = \log|x-1| + \log|C|$

$\log|y+1| = \log|C(x-1)|$

Exponentiate both sides:

This implies $y+1 = C(x-1)$, where $C$ is an arbitrary constant (including 0 and accounting for signs). The general solution is $y = C(x-1) - 1$.

Now, apply the initial condition $y(1) = 2$. Substitute $x=1$ and $y=2$ into the general solution:

$2 = C(0) - 1$

$2 = 0 - 1$

$2 = -1$

This is a contradiction. The equation $2 = -1$ is false, and there is no value of the constant $C$ that can satisfy this equation.

This indicates that there is no solution curve of the form $y = C(x-1) - 1$ that passes through the point $(1, 2)$.

Since the function $f(x,y) = \frac{y+1}{x-1}$ is undefined at the point $(1,2)$ where the initial condition is given, there is no solution to the differential equation that passes through this specific point.

Therefore, the number of solutions of the given differential equation satisfying the initial condition $y(1) = 2$ is zero.

The correct option is (A) none.

The order of a differential equation is the order of the highest derivative present in the equation.

We are given four options and need to identify the second-order differential equation.

Let's examine each option:

(A) $(y')^2 + x = y^2$

This equation involves $y' = \frac{dy}{dx}$. The highest order derivative is the first derivative ($y'$).

Therefore, the order of this differential equation is 1.

(B) $y'' + y = \sin x$

This equation involves $y'' = \frac{d^2y}{dx^2}$ and $y$. The highest order derivative is the second derivative ($y''$).

Therefore, the order of this differential equation is 2.

(C) $y' + (y')^2 + y = 0$

This equation involves $y' = \frac{dy}{dx}$ and $y$. The highest order derivative is the first derivative ($y'$).

Therefore, the order of this differential equation is 1.

(D) $y' = y^2$

This equation involves $y' = \frac{dy}{dx}$ and $y$. The highest order derivative is the first derivative ($y'$).

Therefore, the order of this differential equation is 1.

Comparing the orders of the differential equations in the options, we see that only option (B) is a second-order differential equation.

The correct option is (B) $y'' + y = \sin x$.

Given:

The differential equation $(1 – x^2) \frac{dy}{dx} - xy = 1$.

To Find:

The integrating factor of the differential equation.

Solution:

The given differential equation is:

$(1 – x^2) \frac{dy}{dx} - xy = 1$

This is a first-order differential equation. We need to write it in the standard linear form:

$\frac{dy}{dx} + P(x)y = Q(x)$

Divide the given equation by the coefficient of $\frac{dy}{dx}$, which is $(1 - x^2)$, assuming $1 - x^2 \neq 0$ (i.e., $x \neq \pm 1$).

$\frac{(1 – x^2)}{(1 – x^2)} \frac{dy}{dx} - \frac{x}{(1 – x^2)} y = \frac{1}{(1 – x^2)}$

$\frac{dy}{dx} - \frac{x}{1 – x^2} y = \frac{1}{1 – x^2}$

Comparing this with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we identify $P(x)$:

$P(x) = -\frac{x}{1 – x^2}$

$Q(x) = \frac{1}{1 – x^2}$

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

$IF = e^{\int -\frac{x}{1 – x^2} \, dx}$

Now, we evaluate the integral $\int -\frac{x}{1 – x^2} \, dx$.

Let $u = 1 - x^2$. Then the differential $du = \frac{d}{dx}(1 - x^2) \, dx = -2x \, dx$.

So, $-x \, dx = \frac{1}{2} du$.

Substitute these into the integral:

$\int -\frac{x}{1 – x^2} \, dx = \int \frac{1}{1 – x^2} (-x \, dx) = \int \frac{1}{u} \left(\frac{1}{2} du\right) = \frac{1}{2} \int \frac{1}{u} \, du$

$\int -\frac{x}{1 – x^2} \, dx = \frac{1}{2} \ln|u| + C'$

Substitute back $u = 1 - x^2$:

$\int -\frac{x}{1 – x^2} \, dx = \frac{1}{2} \ln|1 – x^2| + C'$

Using the property of logarithms $a \ln b = \ln b^a$ and for the integrating factor, we take the constant of integration $C' = 0$:

$\int -\frac{x}{1 – x^2} \, dx = \ln(|1 – x^2|^{\frac{1}{2}}) = \ln\sqrt{|1 – x^2|}$

Now, substitute this back into the formula for the integrating factor:

$IF = e^{\int P(x) dx} = e^{\ln\sqrt{|1 – x^2|}}$

Using the property $e^{\ln a} = a$ (for $a > 0$):

$IF = \sqrt{|1 – x^2|}$

Since one of the options is $\sqrt{1-x^2}$, this implies we are likely considering the interval where $1 - x^2 > 0$, i.e., $-1 < x < 1$. In this interval, $|1 - x^2| = 1 - x^2$.

So, the integrating factor is $\sqrt{1 – x^2}$.

Comparing this with the given options:

(A) -x

(B) $\frac{x}{1 + x^2}$

(C) $\sqrt{1 − x^2}$

(D) $\frac{1}{2} \log (1 - x^2)$

The calculated integrating factor matches option (C).

The correct option is (C) $\sqrt{1 − x^2}$.

Given:

The general solution of a differential equation is $\tan^{-1} x + \tan^{-1} y = c$, where $c$ is an arbitrary constant.

To Find:

The differential equation corresponding to the given general solution.

Solution:

We are given the relation:

$\tan^{-1} x + \tan^{-1} y = c$

To find the differential equation, we differentiate the given relation with respect to $x$.

$\frac{d}{dx} (\tan^{-1} x + \tan^{-1} y) = \frac{d}{dx} (c)$

Using the chain rule, the derivative of $\tan^{-1} x$ with respect to $x$ is $\frac{1}{1+x^2}$, and the derivative of $\tan^{-1} y$ with respect to $x$ is $\frac{1}{1+y^2} \frac{dy}{dx}$. The derivative of a constant $c$ is 0.

So, differentiating the equation gives:

$\frac{1}{1+x^2} + \frac{1}{1+y^2} \frac{dy}{dx} = 0$

This is the differential equation. We can rearrange it to match the forms given in the options.

We can write $\frac{dy}{dx}$ as $\frac{dy}{dx}$ or implicitly multiply by $dx$ to get differential form.

Multiply the equation by $(1+x^2)(1+y^2)$ to clear denominators (assuming $x^2 \neq -1$ and $y^2 \neq -1$, which is always true for real $x$ and $y$):

$(1+x^2)(1+y^2) \left( \frac{1}{1+x^2} + \frac{1}{1+y^2} \frac{dy}{dx} \right) = (1+x^2)(1+y^2) \cdot 0$

$(1+y^2) + (1+x^2) \frac{dy}{dx} = 0$

Rearranging this equation:

$(1+x^2) \frac{dy}{dx} = -(1+y^2)$

We can also express this in terms of differentials $dx$ and $dy$ by recalling that $\frac{dy}{dx} = \frac{dy}{dx}$ (literally, not in terms of the ratio of infinitesimals, but conceptually leading to the differential form):

Multiply the equation $\frac{1}{1+x^2} + \frac{1}{1+y^2} \frac{dy}{dx} = 0$ by $dx$ (informally, or think of it as rearranging the terms):

$\frac{1}{1+x^2} \, dx + \frac{1}{1+y^2} \, dy = 0$

$(1+y^2) \, dx + (1+x^2) \, dy = 0$

Alternatively, from $(1+y^2) + (1+x^2) \frac{dy}{dx} = 0$, multiply by $dx$: $(1+y^2) dx + (1+x^2) dy = 0$.

Comparing this differential equation with the given options:

(A) $\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}$

(B) $\frac{dy}{dx} = \frac{1 + x^2}{1 + y^2}$

(C) (1 + x²) dy + (1 + y²) dx = 0

(D) (1 + x²) dx + (1 + y²) dy = 0

Our derived equation is $(1+x^2) dy + (1+y^2) dx = 0$, which matches option (C).

The correct option is (C) (1 + x²) dy + (1 + y²) dx = 0.

Given:

The differential equation $y \frac{dy}{dx} + x = c$, where $c$ is a constant.

To Determine:

The geometric shape represented by the solutions of the differential equation.

Solution:

The given differential equation is:

$y \frac{dy}{dx} + x = c$

We can rewrite this equation to separate the variables:

$y \frac{dy}{dx} = c - x$

$y \, dy = (c - x) \, dx$

Now, integrate both sides of the equation:

$\int y \, dy = \int (c - x) \, dx$

Evaluate the integrals:

$\int y \, dy = \frac{y^2}{2} + K_1$

$\int (c - x) \, dx = \int c \, dx - \int x \, dx = cx - \frac{x^2}{2} + K_2$

where $K_1$ and $K_2$ are constants of integration.

Equating the results of the integration:

$\frac{y^2}{2} + K_1 = cx - \frac{x^2}{2} + K_2$

Rearrange the terms to group $x$ and $y$ together and consolidate the constants:

$\frac{x^2}{2} - cx + \frac{y^2}{2} = K_2 - K_1$

Let $K = K_2 - K_1$, which is a new arbitrary constant.

$\frac{x^2}{2} - cx + \frac{y^2}{2} = K$

Multiply the entire equation by 2:

$x^2 - 2cx + y^2 = 2K$

To identify the geometric shape, we complete the square for the $x$ terms. Add $c^2$ to both sides of the equation:

$x^2 - 2cx + c^2 + y^2 = 2K + c^2$

The terms $x^2 - 2cx + c^2$ form a perfect square $(x-c)^2$. Let $R^2 = 2K + c^2$, which is a non-negative constant for a real circle.

$(x - c)^2 + y^2 = R^2$

This equation is in the standard form of a circle with center at $(h, k) = (c, 0)$ and radius $r = \sqrt{R^2}$ (assuming $R^2 > 0$). If $R^2 = 0$, it represents a point $(c, 0)$, which can be considered a circle of zero radius. If $R^2 < 0$, there is no real solution, but the general form describes a family of circles as the integration constant $K$ (and thus $R^2$) varies.

Thus, the solution of the differential equation represents a family of circles with centers on the x-axis at the point $(c, 0)$.

Comparing this with the given options:

(A) Family of hyperbolas

(B) Family of parabolas

(C) Family of ellipses

(D) Family of circles

Our result matches option (D).

The correct option is (D) Family of circles.

Given:

The differential equation $e^x \cos y \, dx – e^x \sin y \, dy = 0$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is in the form $M(x, y) \, dx + N(x, y) \, dy = 0$, where:

$M(x, y) = e^x \cos y$

$N(x, y) = -e^x \sin y$

Let's check if the equation is exact by verifying if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$.

Calculate $\frac{\partial M}{\partial y}$:

$\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(e^x \cos y) = e^x \frac{\partial}{\partial y}(\cos y) = e^x (-\sin y) = -e^x \sin y$

Calculate $\frac{\partial N}{\partial x}$:

$\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(-e^x \sin y) = (-\sin y) \frac{\partial}{\partial x}(e^x) = (-\sin y) (e^x) = -e^x \sin y$

Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} = -e^x \sin y$, the given differential equation is exact.

The general solution of an exact differential equation is given by $\int M(x, y) \, dx + \int (\text{terms in } N(x, y) \text{ not containing } x) \, dy = C$, where $C$ is an arbitrary constant.

Evaluate the first integral $\int M(x, y) \, dx$, treating $y$ as a constant:

$\int e^x \cos y \, dx = \cos y \int e^x \, dx = \cos y \cdot e^x$

Now identify the terms in $N(x, y) = -e^x \sin y$ that do not contain $x$. The entire term $-e^x \sin y$ contains the factor $e^x$, which depends on $x$. Therefore, there are no terms in $N(x, y)$ that are purely functions of $y$. The second integral part of the general solution formula is $\int 0 \, dy = 0$.

The general solution is thus obtained by setting the result of the first integral plus the result of the second integral equal to a constant $C$ (or $k$ as used in the options):

$\int e^x \cos y \, dx + \int 0 \, dy = k$

$e^x \cos y = k$

This is the general solution of the given differential equation.

Comparing this solution with the given options:

(A) $e^x \cos y = k$

(B) $e^x \sin y = k$

(C) $e^x = k \cos y$

(D) $e^x = k \sin y$

Our derived solution matches option (A).

The correct option is (A) $e^x \cos y = k$.

Given:

The differential equation $\frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^3 + 6y^5 = 0$.

To Find:

The degree of the differential equation.

Solution:

The given differential equation is:

$\frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^3 + 6y^5 = 0$

To determine the degree of a differential equation, we first identify the highest order derivative present in the equation. The order of the differential equation is the order of this highest derivative.

In the given equation, the derivatives present are $\frac{d^2y}{dx^2}$ and $\frac{dy}{dx}$.

The highest order derivative is $\frac{d^2y}{dx^2}$, which is a second-order derivative. Thus, the order of the differential equation is 2.

The degree of a differential equation is the highest power of the highest order derivative when the equation is expressed as a polynomial in terms of its derivatives. For the degree to be defined, the differential equation must be a polynomial equation in its derivatives.

The given equation $\frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^3 + 6y^5 = 0$ is already a polynomial in terms of the derivatives $\frac{d^2y}{dx^2}$ and $\frac{dy}{dx}$, as well as $y$.

The highest order derivative is $\frac{d^2y}{dx^2}$. Its power in the equation is 1.

The highest power of the highest order derivative ($\frac{d^2y}{dx^2}$) is 1.

Therefore, the degree of the differential equation is 1.

Comparing this with the given options:

(A) 1

(B) 2

(C) 3

(D) 5

The calculated degree matches option (A).

The correct option is (A) 1.

Given:

The differential equation $\frac{dy}{dx} + y = e^{-x}$ and the initial condition $y(0) = 0$.

To Find:

The particular solution of the differential equation satisfying the initial condition.

Solution:

The given differential equation is:

$\frac{dy}{dx} + y = e^{-x}$

This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.

Comparing the given equation with the standard form, we have:

$P(x) = 1$

$Q(x) = e^{-x}$

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

$IF = e^{\int 1 \, dx}$

$IF = e^x$

The general solution of a first-order linear differential equation is given by:

$y \cdot (IF) = \int Q(x) \cdot (IF) \, dx + C$

Substitute the values of $y$, $IF$, and $Q(x)$:

$y \cdot e^x = \int e^{-x} \cdot e^x \, dx + C$

$y e^x = \int e^{(-x+x)} \, dx + C$

$y e^x = \int e^0 \, dx + C$

$y e^x = \int 1 \, dx + C$

$y e^x = x + C$

The general solution is $y e^x = x + C$. We can write this explicitly for $y$:

$y = \frac{x+C}{e^x}$

$y = (x+C)e^{-x}$

Now, we use the initial condition $y(0) = 0$ to find the value of the arbitrary constant C.

Substitute $x=0$ and $y=0$ into the general solution:

$0 = (0 + C)e^{-0}$

$0 = C \cdot e^0$

$0 = C \cdot 1$

$C = 0$

Substitute the value of C back into the general solution to obtain the particular solution:

$y = (x + 0)e^{-x}$

$y = x e^{-x}$

Comparing this particular solution with the given options:

(A) y = e^x (x – 1)

(B) y = x e^–x

(C) y = x e^–x + 1

(D) y = (x + 1)e^–x

Our derived particular solution matches option (B).

The correct option is (B) y = x e^–x.

Given:

The differential equation $\frac{dy}{dx} + y \tan x - \sec x = 0$.

To Find:

The integrating factor of the differential equation.

Solution:

The given differential equation is:

$\frac{dy}{dx} + y \tan x - \sec x = 0$

Rewrite the equation in the standard form of a first-order linear differential equation:

$\frac{dy}{dx} + P(x)y = Q(x)$

$\frac{dy}{dx} + (\tan x)y = \sec x$

Comparing this equation with the standard form, we identify $P(x)$:

$P(x) = \tan x$

$Q(x) = \sec x$

The integrating factor (IF) is given by the formula:

$IF = e^{\int P(x) dx}$

Substitute $P(x) = \tan x$ into the formula:

$IF = e^{\int \tan x \, dx}$

Now, evaluate the integral $\int \tan x \, dx$:

$\int \tan x \, dx = \int \frac{\sin x}{\cos x} \, dx$

Let $u = \cos x$. Then $du = -\sin x \, dx$, so $\sin x \, dx = -du$.

$\int \frac{\sin x}{\cos x} \, dx = \int \frac{-du}{u} = -\int \frac{1}{u} \, du = -\ln|u|$

Substitute back $u = \cos x$:

$\int \tan x \, dx = -\ln|\cos x|$

Using logarithm properties, $-\ln|\cos x| = \ln(|\cos x|^{-1}) = \ln|\frac{1}{\cos x}| = \ln|\sec x|$.

For the integrating factor, we can use $\ln|\sec x|$ or $\ln(\sec x)$ assuming $\sec x > 0$ in the interval of interest.

Substitute the result of the integral back into the IF formula:

$IF = e^{\ln|\sec x|}$

Using the property $e^{\ln a} = a$ for $a>0$. Assuming $\sec x > 0$:

$IF = \sec x$

The integrating factor of the given differential equation is $\sec x$.

Comparing this with the given options:

(A) cos x

(B) sec x

(C) e^{cos x}

(D) e^{sec x}

The calculated integrating factor matches option (B).

The correct option is (B) sec x.

Given:

The differential equation $\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is:

$\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}$

This is a separable differential equation. We can separate the variables $x$ and $y$ by rearranging the terms:

$\frac{dy}{1 + y^2} = \frac{dx}{1 + x^2}$

Now, integrate both sides of the equation:

$\int \frac{dy}{1 + y^2} = \int \frac{dx}{1 + x^2}$

The integral of $\frac{1}{1+t^2}$ with respect to $t$ is $\tan^{-1} t + C$. Applying this formula to both sides, we get:

$\tan^{-1} y = \tan^{-1} x + C$

where $C$ is the constant of integration.

Rearrange the equation to gather the inverse tangent terms:

$\tan^{-1} y - \tan^{-1} x = C$

We can use the inverse tangent identity $\tan^{-1} A - \tan^{-1} B = \tan^{-1} \left( \frac{A - B}{1 + AB} \right)$, provided $AB > -1$. Let $A=y$ and $B=x$.

$\tan^{-1} \left( \frac{y - x}{1 + yx} \right) = C$

Taking the tangent of both sides:

$\frac{y - x}{1 + xy} = \tan(C)$

Let $k = \tan(C)$. Since $C$ is an arbitrary constant, $\tan(C)$ is also an arbitrary constant (it can take any real value). So, let $k$ be the arbitrary constant.

$\frac{y - x}{1 + xy} = k$

Multiply both sides by $(1 + xy)$ to get the solution in a different form:

$y - x = k(1 + xy)$

This is the general solution of the given differential equation.

Comparing this solution with the given options:

(A) y = tan^–1 x (This is only a specific case if $C=0$ and $x=0$. Not the general solution.)

(B) y – x = k (1 + xy)

(C) x = tan^–1 y (This corresponds to $\tan^{-1} x - \tan^{-1} y = C$, which is slightly different in the sign of k.)

(D) tan (xy) = k (This is not the correct form of the solution).

Our derived solution matches option (B).

The correct option is (B) y – x = k (1 + xy).

Given:

The differential equation $\frac{dy}{dx} + y = \frac{1 + y}{x}$.

To Find:

The integrating factor of the differential equation.

Solution:

The given differential equation is:

$\frac{dy}{dx} + y = \frac{1 + y}{x}$

To find the integrating factor, we need to write this equation in the standard form of a first-order linear differential equation:

$\frac{dy}{dx} + P(x)y = Q(x)$

Rearrange the terms in the given equation:

$\frac{dy}{dx} + y = \frac{1}{x} + \frac{y}{x}$

Move the term involving $y$ from the right side to the left side:

$\frac{dy}{dx} + y - \frac{y}{x} = \frac{1}{x}$

Factor out $y$ from the terms on the left side:

$\frac{dy}{dx} + y \left( 1 - \frac{1}{x} \right) = \frac{1}{x}$

$\frac{dy}{dx} + \left( \frac{x - 1}{x} \right) y = \frac{1}{x}$

Now, comparing this equation with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we identify $P(x)$ and $Q(x)$:

$P(x) = \frac{x - 1}{x} = 1 - \frac{1}{x}$

$Q(x) = \frac{1}{x}$

The integrating factor (IF) is given by the formula:

$IF = e^{\int P(x) dx}$

Substitute $P(x) = 1 - \frac{1}{x}$ into the formula:

$IF = e^{\int \left( 1 - \frac{1}{x} \right) dx}$

Now, evaluate the integral $\int \left( 1 - \frac{1}{x} \right) dx$:

$\int \left( 1 - \frac{1}{x} \right) dx = \int 1 \, dx - \int \frac{1}{x} \, dx = x - \ln|x| + C'$

For the integrating factor, we can take the constant of integration $C'$ to be 0.

$\int \left( 1 - \frac{1}{x} \right) dx = x - \ln|x|$

Substitute this result back into the formula for the integrating factor:

$IF = e^{x - \ln|x|}$

Using the properties of exponents and logarithms ($e^{a-b} = e^a e^{-b}$ and $e^{- \ln c} = e^{\ln c^{-1}} = c^{-1} = \frac{1}{c}$):

$IF = e^x \cdot e^{-\ln|x|}$

$IF = e^x \cdot \frac{1}{|x|}$

Assuming we are working on an interval where $x > 0$, $|x| = x$. The integrating factor is:

$IF = e^x \cdot \frac{1}{x} = \frac{e^x}{x}$

The integrating factor of the given differential equation is $\frac{e^x}{x}$.

Comparing this with the given options:

(A) $\frac{x}{e^x}$

(B) $\frac{e^x}{x}$

(C) xe^x

(D) e^x

The calculated integrating factor matches option (B).

The correct option is (B) $\frac{e^x}{x}$.

Given:

The function $y = ae^{mx} + be^{-mx}$, where $a$ and $b$ are arbitrary constants.

To Find:

The differential equation satisfied by the given function.

Solution:

We are given the function:

$y = ae^{mx} + be^{-mx}$

Since there are two arbitrary constants ($a$ and $b$), we expect the corresponding differential equation to be of order 2. We need to find the first and second derivatives of $y$ with respect to $x$.

Calculate the first derivative $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{d}{dx}(ae^{mx} + be^{-mx})$

$\frac{dy}{dx} = a \frac{d}{dx}(e^{mx}) + b \frac{d}{dx}(e^{-mx})$

Using the chain rule, $\frac{d}{dx}(e^{kx}) = ke^{kx}$.

$\frac{dy}{dx} = a(me^{mx}) + b(-me^{-mx})$

Calculate the second derivative $\frac{d^2y}{dx^2}$ by differentiating equation (i) with respect to $x$:

$\frac{d^2y}{dx^2} = \frac{d}{dx}(ame^{mx} - bme^{-mx})$

$\frac{d^2y}{dx^2} = am \frac{d}{dx}(e^{mx}) - bm \frac{d}{dx}(e^{-mx})$

$\frac{d^2y}{dx^2} = am(me^{mx}) - bm(-me^{-mx})$

$\frac{d^2y}{dx^2} = a m^2 e^{mx} + b m^2 e^{-mx}$

Factor out $m^2$ from the expression for $\frac{d^2y}{dx^2}$:

$\frac{d^2y}{dx^2} = m^2 (ae^{mx} + be^{-mx})$

Observe that the term in the parenthesis is the original function $y$. Substitute $y$ back into the equation:

$\frac{d^2y}{dx^2} = m^2 y$

Rearrange the equation to form the differential equation:

$\frac{d^2y}{dx^2} - m^2 y = 0$

This is the required differential equation satisfied by the given function.

Comparing this equation with the given options:

(A) $\frac{dy}{dx} + my = 0$

(B) $\frac{dy}{dx} - my = 0$

(C) $\frac{d^2y}{dx^2} - m^2y = 0$

(D) $\frac{d^2y}{dx^2} + m^2y = 0$

The derived equation matches option (C).

The correct option is (C) $\frac{d^2y}{dx^2} - m^2y = 0$.

Given:

The differential equation $\cos x \sin y \, dx + \sin x \cos y \, dy = 0$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is:

$\cos x \sin y \, dx + \sin x \cos y \, dy = 0$

This is a separable differential equation. To separate the variables $x$ and $y$, divide the entire equation by $(\sin x \sin y)$, assuming $\sin x \neq 0$ and $\sin y \neq 0$.

$\frac{\cos x \sin y}{\sin x \sin y} \, dx + \frac{\sin x \cos y}{\sin x \sin y} \, dy = \frac{0}{\sin x \sin y}$

$\frac{\cos x}{\sin x} \, dx + \frac{\cos y}{\sin y} \, dy = 0$

This can be written in terms of cotangent functions:

$\cot x \, dx + \cot y \, dy = 0$

Now, integrate both sides of the separated equation:

$\int \cot x \, dx + \int \cot y \, dy = \int 0 \, dx$

Evaluate the integrals:

The integral of $\cot u$ with respect to $u$ is $\log|\sin u|$ (using $\log$ for natural logarithm as requested).

$\int \cot x \, dx = \log|\sin x| + C_1$

$\int \cot y \, dy = \log|\sin y| + C_2$

The integral of 0 is a constant, let's call it $C_3$.

Combining the integrated terms:

$\log|\sin x| + \log|\sin y| = C_3 - C_1 - C_2$

Let $C = C_3 - C_1 - C_2$, which is an arbitrary constant.

$\log|\sin x| + \log|\sin y| = C$

Using the property of logarithms, $\log a + \log b = \log (ab)$:

$\log|\sin x \sin y| = C$

Exponentiate both sides (using $e^C$ as a new constant):

$|\sin x \sin y| = e^C$

Let $k = \pm e^C$. Since $C$ is an arbitrary constant, $e^C$ is a positive constant, and $\pm e^C$ is an arbitrary non-zero constant. If we consider the cases where $\sin x = 0$ or $\sin y = 0$, these also satisfy the original equation (0=0), and correspond to $k=0$. Thus, $k$ can be any arbitrary constant.

$\sin x \sin y = k$

This is the general solution of the given differential equation.

Comparing this solution with the given options:

(A) $\frac{\sin x}{\sin y} = c$ (equivalent to $\sin x = c \sin y$)

(B) $\sin x \; \sin y = c$

(C) $\sin x + \sin y = c$

(D) $\cos x \; \cos y = c$

Our derived solution matches option (B).

The correct option is (B) $\sin x \; \sin y = c$.

Given:

The differential equation $x \frac{dy}{dx} + y = e^x$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is:

$x \frac{dy}{dx} + y = e^x$

This is a first-order linear differential equation. We need to write it in the standard form:

$\frac{dy}{dx} + P(x)y = Q(x)$

Divide the given equation by the coefficient of $\frac{dy}{dx}$, which is $x$ (assuming $x \neq 0$):

$\frac{x}{x} \frac{dy}{dx} + \frac{y}{x} = \frac{e^x}{x}$

$\frac{dy}{dx} + \frac{1}{x} y = \frac{e^x}{x}$

Comparing this with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we identify $P(x)$ and $Q(x)$:

$P(x) = \frac{1}{x}$

$Q(x) = \frac{e^x}{x}$

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

$IF = e^{\int \frac{1}{x} \, dx}$

Evaluate the integral $\int \frac{1}{x} \, dx$. Using log for natural logarithm as requested:

$\int \frac{1}{x} \, dx = \log|x| + C'$

For the integrating factor, we can take the constant of integration $C'$ to be 0. Assuming $x>0$, we use $\log x$.

$\int \frac{1}{x} \, dx = \log x$ (for $x>0$)

Substitute this result back into the formula for the integrating factor:

$IF = e^{\log x}$

Using the property $e^{\log a} = a$ (for $a>0$):

$IF = x$

The general solution of a first-order linear differential equation is given by:

$y \cdot (IF) = \int Q(x) \cdot (IF) \, dx + C$

where $C$ (or $k$ in the options) is the arbitrary constant of integration.

Substitute the values of $y$, $IF$, and $Q(x)$:

$y \cdot x = \int \frac{e^x}{x} \cdot x \, dx + k$

$xy = \int e^x \, dx + k$

Evaluate the integral $\int e^x \, dx$:

$\int e^x \, dx = e^x$

So,

$xy = e^x + k$

To express the solution explicitly for $y$, divide both sides by $x$ (assuming $x \neq 0$):

$y = \frac{e^x + k}{x}$

$y = \frac{e^x}{x} + \frac{k}{x}$

Comparing this general solution with the given options:

(A) $y = \frac{e^x}{x} + \frac{k}{x}$

(B) y = xe^x + cx (Incorrect form)

(C) y = xe^x + k (Incorrect form)

(D) $x = \frac{e^y}{y} + \frac{k}{y}$ (This is the solution for a different type of equation where y is the independent variable)

Our derived solution matches option (A).

The correct option is (A) $y = \frac{e^x}{x} + \frac{k}{x}$.

Given:

The family of curves is given by the equation:

$x^2 + y^2 - 2ay = 0$

where $a$ is the arbitrary constant.

To Find:

The differential equation representing this family of curves.

Solution:

The given equation contains one arbitrary constant $a$. Therefore, the order of the differential equation will be 1.

To obtain the differential equation, we need to eliminate the arbitrary constant $a$ by differentiating the given equation with respect to $x$.

Differentiating $x^2 + y^2 - 2ay = 0$ with respect to $x$:

$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) - \frac{d}{dx}(2ay) = \frac{d}{dx}(0)$

$2x + 2y \frac{dy}{dx} - 2a \frac{dy}{dx} = 0$

$2x + 2(y - a) \frac{dy}{dx} = 0$

Divide by 2:

Now, we need to eliminate the constant $a$ from equation (i) using the original equation. From the original equation $x^2 + y^2 - 2ay = 0$, we can solve for $a$ (assuming $y \neq 0$):

$2ay = x^2 + y^2$

Substitute the value of $a$ from equation (ii) into equation (i):

$x + \left( y - \frac{x^2 + y^2}{2y} \right) \frac{dy}{dx} = 0$

Simplify the term in the parenthesis:

$y - \frac{x^2 + y^2}{2y} = \frac{2y \cdot y - (x^2 + y^2)}{2y} = \frac{2y^2 - x^2 - y^2}{2y} = \frac{y^2 - x^2}{2y}$

Substitute this back into the equation:

$x + \left( \frac{y^2 - x^2}{2y} \right) \frac{dy}{dx} = 0$

Rearrange the equation to get the differential equation:

$\left( \frac{y^2 - x^2}{2y} \right) \frac{dy}{dx} = -x$

$(y^2 - x^2) \frac{dy}{dx} = -x (2y)$

$(y^2 - x^2) \frac{dy}{dx} = -2xy$

Multiply by -1 to match the options:

$-(y^2 - x^2) \frac{dy}{dx} = 2xy$

$(x^2 - y^2) \frac{dy}{dx} = 2xy$

This is the differential equation for the family of curves.

Comparing this differential equation with the given options:

(A) $(x^2 – y^2) \frac{dy}{dx} = 2xy$

(B) $2 (x^2 + y^2) \frac{dy}{dx} = xy$

(C) $2 (x^2 – y^2) \frac{dy}{dx} = xy$

(D) $(x^2 + y^2) \frac{dy}{dx} = 2xy$

Our derived equation matches option (A).

The correct option is (A) $(x^2 – y^2) \frac{dy}{dx} = 2xy$.

Given:

The family of curves is given by the equation:

$y = Ax + A^3$

where A is an arbitrary constant.

To Find:

The order of the differential equation representing this family of curves.

Solution:

The order of the differential equation representing a family of curves is equal to the number of essential arbitrary constants in the equation of the family.

The given equation $y = Ax + A^3$ contains only one arbitrary constant, A.

Therefore, the differential equation representing this family of curves will have an order of 1.

To confirm this, we differentiate the equation with respect to $x$ to eliminate the constant A:

$\frac{dy}{dx} = \frac{d}{dx}(Ax + A^3)$

$\frac{dy}{dx} = A \frac{d}{dx}(x) + \frac{d}{dx}(A^3)$

$\frac{dy}{dx} = A(1) + 0$

$\frac{dy}{dx} = A$

Substituting $A = \frac{dy}{dx}$ back into the original equation $y = Ax + A^3$, we get the differential equation:

$y = x \left( \frac{dy}{dx} \right) + \left( \frac{dy}{dx} \right)^3$

In this differential equation, the highest order derivative is $\frac{dy}{dx}$, which is a first-order derivative. The order of the differential equation is the order of the highest derivative present.

Thus, the order of the differential equation is 1.

Comparing this with the given options:

(A) 3

(B) 2

(C) 1

(D) not defined

Our result matches option (C).

The correct option is (C) 1.

Given:

The differential equation $\frac{dy}{dx} = 2x e^{x^2−y}$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is:

$\frac{dy}{dx} = 2x e^{x^2−y}$

Using the property of exponents $e^{a-b} = e^a e^{-b}$, we can separate the terms involving $x$ and $y$ on the right side:

$\frac{dy}{dx} = 2x e^{x^2} e^{-y}$

This is a separable differential equation. We can separate the variables $x$ and $y$ by rearranging the terms. Multiply both sides by $e^y$ and by $dx$:

$e^y \, dy = 2x e^{x^2} \, dx$

Now, integrate both sides of the separated equation:

$\int e^y \, dy = \int 2x e^{x^2} \, dx$

Evaluate the integral on the left side:

$\int e^y \, dy = e^y$

Evaluate the integral on the right side $\int 2x e^{x^2} \, dx$. Use a substitution.

Let $u = x^2$. Then the differential $du = \frac{d}{dx}(x^2) \, dx = 2x \, dx$.

The integral becomes $\int e^u \, du = e^u$. Substitute back $u = x^2$.

$\int 2x e^{x^2} \, dx = e^{x^2}$

Combine the results of the integration and add the constant of integration, $c$:

$e^y = e^{x^2} + c$

This is the general solution of the given differential equation.

Comparing this solution with the given options:

(A) $e^{x^2−y} = c$ (Equivalent to $e^{x^2}/e^y = c$, or $e^{x^2} = c e^y$. This is not the same form.)

(B) $e^{−y} + e^{x^2} = c$ (Equivalent to $e^{x^2} = c - e^{-y}$, or $e^{x^2} = c - \frac{1}{e^y}$. This is not the same form.)

(C) $e^y = e^{x^2} + c$

(D) $e^{x^2+y} = c$ (Equivalent to $e^{x^2} e^y = c$. Taking log on both sides gives $x^2 + y = \log c$. This is not the same form.)

Our derived solution matches option (C).

The correct option is (C) $e^y = e^{x^2} + c$.

Given:

The slope of the tangent at any point $(x, y)$ on the curve is equal to the ratio of the abscissa ($x$) to the ordinate ($y$) of the point.

To Determine:

The type of curve that satisfies this condition.

Solution:

The slope of the tangent to a curve at a point $(x, y)$ is given by the derivative $\frac{dy}{dx}$.

According to the given condition, the slope of the tangent at $(x, y)$ is equal to the ratio of the abscissa to the ordinate. The abscissa is $x$ and the ordinate is $y$. So, the ratio is $\frac{x}{y}$.

Thus, we have the differential equation:

$\frac{dy}{dx} = \frac{x}{y}$

This is a separable differential equation. Separate the variables:

$y \, dy = x \, dx$

Now, integrate both sides of the equation:

$\int y \, dy = \int x \, dx$

Evaluate the integrals:

$\int y \, dy = \frac{y^2}{2} + C_1$

$\int x \, dx = \frac{x^2}{2} + C_2$

where $C_1$ and $C_2$ are constants of integration.

Equating the results of the integration:

$\frac{y^2}{2} = \frac{x^2}{2} + C$

where $C = C_2 - C_1$ is a new arbitrary constant.

Rearrange the terms:

$\frac{y^2}{2} - \frac{x^2}{2} = C$

Multiply by 2:

$y^2 - x^2 = 2C$

Let $K = 2C$. The equation is $y^2 - x^2 = K$. This can also be written as $x^2 - y^2 = -K$. Let the arbitrary constant be denoted by $c$, so $x^2 - y^2 = c$. This form is more standard for comparison with conic sections.

The general equation for a conic section is $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$.

The equation $x^2 - y^2 = c$ can be written as $x^2 - y^2 - c = 0$. Comparing this with the general form, we have $A=1$, $B=0$, $C=-1$, $D=0$, $E=0$, $F=-c$.

For a conic section without the $xy$ term (i.e., $B=0$), the type of conic is determined by the signs of $A$ and $C$:

• Ellipse: $A$ and $C$ have the same sign (and are non-zero).
• Parabola: Either $A=0$ or $C=0$ (but not both).
• Hyperbola: $A$ and $C$ have opposite signs (and are non-zero).

In the equation $x^2 - y^2 = c$, $A=1$ and $C=-1$. The coefficients $A$ and $C$ have opposite signs and are non-zero. Therefore, the curve is a hyperbola.

Specifically, if $c \neq 0$, it's a hyperbola. If $c=0$, the equation is $x^2 - y^2 = 0$, which factors as $(x-y)(x+y) = 0$. This represents two straight lines $y=x$ and $y=-x$ passing through the origin. These are the asymptotes of the hyperbola family.

The term "rectangular hyperbola" refers to a hyperbola whose asymptotes are perpendicular, which occurs when the equation is of the form $x^2 - y^2 = c$ or $xy = c$. The equation $x^2 - y^2 = c$ indeed represents a rectangular hyperbola (where the angle between asymptotes is $90^\circ$).

Comparing this with the given options:

(A) an ellipse

(B) parabola

(C) circle

(D) rectangular hyperbola

Our result matches option (D).

The correct option is (D) rectangular hyperbola.

Given:

The differential equation $\frac{dy}{dx} = e^{\frac{x^2}{2}} + xy$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is:

$\frac{dy}{dx} = e^{\frac{x^2}{2}} + xy$

Rearrange the equation to put it in the standard form of a first-order linear differential equation:

$\frac{dy}{dx} - xy = e^{\frac{x^2}{2}}$

The standard form is $\frac{dy}{dx} + P(x)y = Q(x)$.

Comparing, we identify $P(x)$ and $Q(x)$:

$P(x) = -x$

$Q(x) = e^{\frac{x^2}{2}}$

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

$IF = e^{\int (-x) \, dx}$

Evaluate the integral $\int (-x) \, dx$:

$\int (-x) \, dx = -\int x \, dx = -\frac{x^2}{2} + C'$

For the integrating factor, we take the constant of integration $C'$ to be 0.

$\int (-x) \, dx = -\frac{x^2}{2}$

Substitute this result back into the formula for the integrating factor:

$IF = e^{-\frac{x^2}{2}}$

The general solution of a first-order linear differential equation is given by:

$y \cdot (IF) = \int Q(x) \cdot (IF) \, dx + C$

where $C$ (or $c$ in the options) is the arbitrary constant of integration.

Substitute the values of $y$, $IF$, and $Q(x)$:

$y \cdot e^{-\frac{x^2}{2}} = \int e^{\frac{x^2}{2}} \cdot e^{-\frac{x^2}{2}} \, dx + c$

$y e^{-\frac{x^2}{2}} = \int e^{\left(\frac{x^2}{2} - \frac{x^2}{2}\right)} \, dx + c$

$y e^{-\frac{x^2}{2}} = \int e^0 \, dx + c$

$y e^{-\frac{x^2}{2}} = \int 1 \, dx + c$

$y e^{-\frac{x^2}{2}} = x + c$

To express the solution explicitly for $y$, multiply both sides by $e^{\frac{x^2}{2}}$:

$y = \frac{x + c}{e^{-\frac{x^2}{2}}}$

$y = (x + c)e^{\frac{x^2}{2}}$

This is the general solution of the given differential equation.

Comparing this general solution with the given options:

(A) $y = ce^{\frac{−x^2}{2}}$

(B) $y = ce^{\frac{x^2}{2}}$

(C) $y = (x + c)e^{\frac{x^2}{2}}$

(D) $y = (c - x)e^{\frac{x^2}{2}}$

Our derived solution matches option (C).

The correct option is (C) $y = (x + c)e^{\frac{x^2}{2}}$.

Given:

The differential equation $(2y – 1) dx – (2x + 3) dy = 0$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is:

$(2y – 1) dx – (2x + 3) dy = 0$

This is a separable differential equation. To separate the variables $x$ and $y$, move the term involving $dy$ to the right side:

$(2y – 1) dx = (2x + 3) dy$

Divide both sides by $(2y - 1)(2x + 3)$, assuming $2y - 1 \neq 0$ and $2x + 3 \neq 0$:

$\frac{(2y – 1)}{(2y – 1)(2x + 3)} \, dx = \frac{(2x + 3)}{(2y – 1)(2x + 3)} \, dy$

$\frac{dx}{2x + 3} = \frac{dy}{2y - 1}$

Now, integrate both sides of the separated equation:

$\int \frac{1}{2x + 3} \, dx = \int \frac{1}{2y - 1} \, dy$

Evaluate the integral on the left side $\int \frac{1}{2x + 3} \, dx$.

Let $u = 2x + 3$. Then $du = 2 \, dx$, so $dx = \frac{1}{2} du$.

$\int \frac{1}{u} \left(\frac{1}{2} du\right) = \frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \log|u|$

Substitute back $u = 2x + 3$:

$\int \frac{1}{2x + 3} \, dx = \frac{1}{2} \log|2x + 3|$

Evaluate the integral on the right side $\int \frac{1}{2y - 1} \, dy$.

Let $v = 2y - 1$. Then $dv = 2 \, dy$, so $dy = \frac{1}{2} dv$.

$\int \frac{1}{v} \left(\frac{1}{2} dv\right) = \frac{1}{2} \int \frac{1}{v} \, dv = \frac{1}{2} \log|v|$

Substitute back $v = 2y - 1$:

$\int \frac{1}{2y - 1} \, dy = \frac{1}{2} \log|2y - 1|$

Combine the results of the integration and add the constant of integration $C$:

$\frac{1}{2} \log|2x + 3| = \frac{1}{2} \log|2y - 1| + C$

Multiply the entire equation by 2:

$\log|2x + 3| = \log|2y - 1| + 2C$

Rearrange the terms:

$\log|2x + 3| - \log|2y - 1| = 2C$

Using the property of logarithms $\log a - \log b = \log (a/b)$:

$\log\left|\frac{2x + 3}{2y - 1}\right| = 2C$

Exponentiate both sides:

$\left|\frac{2x + 3}{2y - 1}\right| = e^{2C}$

Let $k = \pm e^{2C}$. Since $C$ is an arbitrary constant, $e^{2C}$ is a positive constant, and $\pm e^{2C}$ is an arbitrary non-zero constant. We also need to consider the cases where $2x+3=0$ or $2y-1=0$. If $2x+3=0$, the original equation gives $(2y-1) dx - 0 = 0$, so $(2y-1) dx = 0$, which means $2y-1=0$ or $dx=0$. If $2y-1=0$, the original equation gives $0 - (2x+3) dy = 0$, so $(2x+3) dy = 0$, which means $2x+3=0$ or $dy=0$. These cases correspond to $k=0$ or $k=\infty$. Thus, $k$ can be any arbitrary constant.

$\frac{2x + 3}{2y - 1} = k$

This is the general solution of the given differential equation.

Comparing this solution with the given options:

(A) $\frac{2x − 1}{2y + 3} = k$

(B) $\frac{2y + 1}{2x − 3} = k$

(C) $\frac{2x + 3}{2y − 1} = k$

(D) $\frac{2x − 1}{2y − 1} = k$

Our derived solution matches option (C).

The correct option is (C) $\frac{2x + 3}{2y − 1} = k$.

Given:

The function $y = a \cos x + b \sin x$, where $a$ and $b$ are arbitrary constants.

To Find:

The differential equation satisfied by the given function.

Solution:

We are given the function:

$y = a \cos x + b \sin x$

Since there are two arbitrary constants ($a$ and $b$), the differential equation will be of order 2. We need to find the first and second derivatives of $y$ with respect to $x$ to eliminate these constants.

Calculate the first derivative $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{d}{dx}(a \cos x + b \sin x)$

$\frac{dy}{dx} = a \frac{d}{dx}(\cos x) + b \frac{d}{dx}(\sin x)$

$\frac{dy}{dx} = a (-\sin x) + b (\cos x)$

Calculate the second derivative $\frac{d^2y}{dx^2}$ by differentiating equation (i) with respect to $x$:

$\frac{d^2y}{dx^2} = \frac{d}{dx}(-a \sin x + b \cos x)$

$\frac{d^2y}{dx^2} = -a \frac{d}{dx}(\sin x) + b \frac{d}{dx}(\cos x)$

$\frac{d^2y}{dx^2} = -a (\cos x) + b (-\sin x)$

$\frac{d^2y}{dx^2} = -a \cos x - b \sin x$

Factor out $-1$ from the expression for $\frac{d^2y}{dx^2}$:

$\frac{d^2y}{dx^2} = -(a \cos x + b \sin x)$

Observe that the term in the parenthesis is the original function $y$. Substitute $y$ back into the equation:

$\frac{d^2y}{dx^2} = -y$

Rearrange the equation to form the differential equation:

$\frac{d^2y}{dx^2} + y = 0$

This is the required differential equation satisfied by the given function. Note that the differential equation does not contain the arbitrary constants $a$ and $b$. Options (C) and (D) are not valid differential equations in this context as they still contain the arbitrary constants.

Comparing our derived equation with the given options:

(A) $\frac{d^2y}{dx^2} + y = 0$

(B) $\frac{d^2y}{dx^2} - y = 0$

(C) $\frac{d^2y}{dx^2} + (a + b) y = 0$

(D) $\frac{d^2y}{dx^2} + (a - b) y = 0$

The derived equation matches option (A).

The correct option is (A) $\frac{d^2y}{dx^2} + y = 0$.

Given:

The differential equation $\frac{dy}{dx} + y = e^{-x}$ and the initial condition $y(0) = 0$.

To Find:

The particular solution of the differential equation satisfying the initial condition.

Solution:

The given differential equation is:

$\frac{dy}{dx} + y = e^{-x}$

This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.

Comparing the given equation with the standard form, we have:

$P(x) = 1$

$Q(x) = e^{-x}$

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

$IF = e^{\int 1 \, dx}$

$IF = e^x$

The general solution of a first-order linear differential equation is given by:

$y \cdot (IF) = \int Q(x) \cdot (IF) \, dx + C$

where $C$ is the arbitrary constant of integration.

Substitute the values of $y$, $IF$, and $Q(x)$:

$y \cdot e^x = \int e^{-x} \cdot e^x \, dx + C$

$y e^x = \int e^{(-x+x)} \, dx + C$

$y e^x = \int e^0 \, dx + C$

$y e^x = \int 1 \, dx + C$

$y e^x = x + C$

The general solution is $y e^x = x + C$. We can write this explicitly for $y$:

$y = \frac{x+C}{e^x}$

$y = (x+C)e^{-x}$

Now, we use the initial condition $y(0) = 0$ to find the value of the arbitrary constant C.

Substitute $x=0$ and $y=0$ into the general solution:

$0 = (0 + C)e^{-0}$

$0 = C \cdot e^0$

$0 = C \cdot 1$

$C = 0$

Substitute the value of C back into the general solution to obtain the particular solution:

$y = (x + 0)e^{-x}$

$y = x e^{-x}$

Comparing this particular solution with the given options:

(A) y = e^–x (x – 1)

(B) y = xe^x

(C) y = xe^–x + 1

(D) y = xe^–x

Our derived particular solution matches option (D).

The correct option is (D) y = xe^–x.

Given:

The differential equation $\left( \frac{d^3y}{dx^3} \right)^2− 3\frac{d^2y}{dx^2} + 2 \left( \frac{dy}{dx} \right)^4 = y^4$.

To Find:

The order and degree of the differential equation.

Solution:

The given differential equation is:

$\left( \frac{d^3y}{dx^3} \right)^2− 3\frac{d^2y}{dx^2} + 2 \left( \frac{dy}{dx} \right)^4 = y^4$

The order of a differential equation is the order of the highest derivative present in the equation.

The derivatives present in the equation are $\frac{d^3y}{dx^3}$, $\frac{d^2y}{dx^2}$, and $\frac{dy}{dx}$.

The orders of these derivatives are 3, 2, and 1, respectively.

The highest order derivative is $\frac{d^3y}{dx^3}$.

Therefore, the order of the differential equation is 3.

The degree of a differential equation is the highest power of the highest order derivative present in the differential equation when it is a polynomial in terms of its derivatives. The given equation is already a polynomial in its derivatives.

The highest order derivative is $\frac{d^3y}{dx^3}$. Its power in the equation is 2 (from the term $\left( \frac{d^3y}{dx^3} \right)^2$).

Although other derivatives like $\frac{dy}{dx}$ have a higher power (4), the degree is determined by the power of the *highest order* derivative.

The highest power of the highest order derivative ($\frac{d^3y}{dx^3}$) is 2.

Therefore, the degree of the differential equation is 2.

The order and degree are 3 and 2, respectively.

Comparing this with the given options:

(A) 1, 4 (Order 1, Degree 4)

(B) 3, 4 (Order 3, Degree 4)

(C) 2, 4 (Order 2, Degree 4)

(D) 3, 2 (Order 3, Degree 2)

Our calculated order (3) and degree (2) match option (D).

The correct option is (D) 3, 2.

Given:

The differential equation $\left[ 1 + \left( \frac{dy}{dx} \right)^2 \right] = \frac{d^2y}{dx^2}$.

To Find:

The order and degree of the differential equation.

Solution:

The given differential equation is:

$\left[ 1 + \left( \frac{dy}{dx} \right)^2 \right] = \frac{d^2y}{dx^2}$

The order of a differential equation is the order of the highest derivative present in the equation.

The derivatives present in the equation are $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$.

The orders of these derivatives are 1 and 2, respectively.

The highest order derivative is $\frac{d^2y}{dx^2}$.

Therefore, the order of the differential equation is 2.

The degree of a differential equation is the highest power of the highest order derivative present in the differential equation when it is a polynomial in terms of its derivatives. The given equation is already a polynomial in its derivatives.

The highest order derivative is $\frac{d^2y}{dx^2}$. The equation is $\frac{d^2y}{dx^2} = 1 + \left( \frac{dy}{dx} \right)^2$.

The power of the highest order derivative ($\frac{d^2y}{dx^2}$) in this equation is 1.

Therefore, the degree of the differential equation is 1.

The order and degree are 2 and 1, respectively.

Comparing this with the given options:

(A) 2, $\frac{3}{2}$ (Order 2, Degree 3/2 - fractional degree implies the equation was not in polynomial form before squaring, as in Question 35. This equation is already in polynomial form.)

(B) 2, 3 (Order 2, Degree 3)

(C) 2, 1 (Order 2, Degree 1)

(D) 3, 4 (Order 3, Degree 4)

Our calculated order (2) and degree (1) match option (C).

The correct option is (C) 2, 1.

Given:

The family of curves is given by the equation:

$y^2 = 4a (x + a)$

$y^2 = 4ax + 4a^2$

where $a$ is the arbitrary constant.

To Find:

The differential equation representing this family of curves.

Solution:

The given equation contains one arbitrary constant $a$. Therefore, the order of the differential equation will be 1.

To obtain the differential equation, we need to eliminate the arbitrary constant $a$. Differentiate the given equation with respect to $x$:

$\frac{d}{dx}(y^2) = \frac{d}{dx}(4ax + 4a^2)$

Using the chain rule on the left side and the derivative rules on the right side (treating $a$ as a constant):

$2y \frac{dy}{dx} = 4a \frac{d}{dx}(x) + \frac{d}{dx}(4a^2)$

$2y \frac{dy}{dx} = 4a(1) + 0$

From equation (i), we can express the arbitrary constant $a$ in terms of $y$ and $\frac{dy}{dx}$ (assuming $y \neq 0$):

$a = \frac{2y \frac{dy}{dx}}{4} = \frac{y}{2} \frac{dy}{dx}$

Now, substitute this expression for $a$ back into the original equation $y^2 = 4a(x+a)$:

$y^2 = 4 \left( \frac{y}{2} \frac{dy}{dx} \right) \left( x + \frac{y}{2} \frac{dy}{dx} \right)$

Simplify the right side:

$y^2 = 2y \frac{dy}{dx} \left( x + \frac{y}{2} \frac{dy}{dx} \right)$

Distribute the term $2y \frac{dy}{dx}$:

$y^2 = \left( 2y \frac{dy}{dx} \right) x + \left( 2y \frac{dy}{dx} \right) \left( \frac{y}{2} \frac{dy}{dx} \right)$

$y^2 = 2xy \frac{dy}{dx} + 2y \cdot \frac{y}{2} \left( \frac{dy}{dx} \right)^2$

$y^2 = 2xy \frac{dy}{dx} + y^2 \left( \frac{dy}{dx} \right)^2$

Rearrange the terms to form a standard differential equation (assuming $y \neq 0$, we can see the terms involving $y$):

$0 = 2xy \frac{dy}{dx} + y^2 \left( \frac{dy}{dx} \right)^2 - y^2$

If $y \neq 0$, we can divide the entire equation by $y$:

$0 = 2x \frac{dy}{dx} + y \left( \frac{dy}{dx} \right)^2 - y$

This is the differential equation representing the family of curves (excluding the case $y=0$, which is a solution if $a=0$).

The equation can be written as:

$2x \frac{dy}{dx} + y \left( \frac{dy}{dx} \right)^2 - y = 0$

Comparing this equation with the given options, option (D) matches the left side of our derived equation. It is implied that the differential equation is formed by setting this expression equal to zero.

The correct option is (D) $2x \frac{dy}{dx} + y \left( \frac{dy}{dx} \right)^2 - y$ (interpreted as equal to zero).

Given:

The differential equation $\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 0$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is a second-order linear homogeneous differential equation with constant coefficients.

The equation is:

$\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 0$

The characteristic equation is obtained by replacing $\frac{d^2y}{dx^2}$ with $m^2$, $\frac{dy}{dx}$ with $m$, and $y$ with 1:

We solve this quadratic equation for $m$. The left side is a perfect square trinomial:

$(m - 1)^2 = 0$

This equation has a repeated real root:

$m = 1$ (with multiplicity 2)

When the characteristic equation has a repeated real root $m$, the general solution of the second-order linear homogeneous differential equation is given by:

$y = (Ax + B)e^{mx}$

where A and B are arbitrary constants.

Substitute the root $m=1$ into the general solution formula:

$y = (Ax + B)e^{1 \cdot x}$

$y = (Ax + B)e^x$

This is the general solution of the given differential equation.

Comparing this general solution with the given options:

(A) y = (Ax + B)e^x

(B) y = (Ax + B)e^–x

(C) y = Ae^x + Be^–x (This form is for distinct real roots)

(D) y = A cos x + B sin x (This form is for complex conjugate roots)

Our derived solution matches option (A).

The correct option is (A) y = (Ax + B)e^x.

Given:

The differential equation $\frac{dy}{dx} + y \tan x = \sec x$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is:

$\frac{dy}{dx} + y \tan x = \sec x$

This is a first-order linear differential equation in the standard form $\frac{dy}{dx} + P(x)y = Q(x)$.

Comparing, we identify $P(x)$ and $Q(x)$:

$P(x) = \tan x$

$Q(x) = \sec x$

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

$IF = e^{\int \tan x \, dx}$

We know that $\int \tan x \, dx = \log|\sec x|$ (using log for natural logarithm as requested).

$IF = e^{\log|\sec x|}$

Using the property $e^{\log a} = a$ (for $a>0$). Assuming $\sec x > 0$ in the interval of interest:

$IF = \sec x$

The general solution of a first-order linear differential equation is given by:

$y \cdot (IF) = \int Q(x) \cdot (IF) \, dx + C$

where $C$ (or $c$ in the options) is the arbitrary constant of integration.

Substitute the values of $y$, $IF$, and $Q(x)$:

$y \cdot \sec x = \int \sec x \cdot \sec x \, dx + c$

$y \sec x = \int \sec^2 x \, dx + c$

Evaluate the integral $\int \sec^2 x \, dx$:

$\int \sec^2 x \, dx = \tan x$

So,

$y \sec x = \tan x + c$

This is the general solution of the given differential equation.

Comparing this general solution with the given options:

(A) y sec x = tan x + c

(B) y tan x = sec x + c

(C) tan x = y tan x + c

(D) x sec x = tan y + c

Our derived solution matches option (A).

The correct option is (A) y sec x = tan x + c.

Given:

The differential equation $\frac{dy}{dx} + \frac{y}{x} = \sin x$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is:

$\frac{dy}{dx} + \frac{1}{x} y = \sin x$

This is a first-order linear differential equation in the standard form $\frac{dy}{dx} + P(x)y = Q(x)$.

Comparing the given equation with the standard form, we identify $P(x)$ and $Q(x)$:

$P(x) = \frac{1}{x}$

$Q(x) = \sin x$

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

$IF = e^{\int \frac{1}{x} \, dx}$

Evaluate the integral $\int \frac{1}{x} \, dx$. Using log for natural logarithm as requested:

$\int \frac{1}{x} \, dx = \log|x| + C'$

For the integrating factor, we can take the constant of integration $C'$ to be 0. Assuming $x>0$, we use $\log x$.

$\int \frac{1}{x} \, dx = \log x$ (for $x>0$)

Substitute this result back into the formula for the integrating factor:

$IF = e^{\log x}$

Using the property $e^{\log a} = a$ (for $a>0$):

$IF = x$

The general solution of a first-order linear differential equation is given by:

$y \cdot (IF) = \int Q(x) \cdot (IF) \, dx + C$

where $C$ (or $c$ in the options) is the arbitrary constant of integration.

Substitute the values of $y$, $IF$, and $Q(x)$:

$y \cdot x = \int \sin x \cdot x \, dx + c$

$xy = \int x \sin x \, dx + c$

Now, evaluate the integral $\int x \sin x \, dx$ using integration by parts ($\int u \, dv = uv - \int v \, du$).

Let $u = x$, so $du = dx$.

Let $dv = \sin x \, dx$, so $v = \int \sin x \, dx = -\cos x$.

$\int x \sin x \, dx = x(-\cos x) - \int (-\cos x) \, dx$

$= -x \cos x + \int \cos x \, dx$

$= -x \cos x + \sin x$

Substitute the result of the integral back into the general solution equation:

$xy = -x \cos x + \sin x + c$

Rearrange the terms to match the form of the options. Move the term $-x \cos x$ to the left side:

$xy + x \cos x = \sin x + c$

Factor out $x$ from the terms on the left side:

$x(y + \cos x) = \sin x + c$

This is the general solution of the given differential equation.

Comparing this general solution with the given options:

(A) x (y + cos x) = sin x + c

(B) x (y – cos x) = sin x + c

(C) xy cos x = sin x + c

(D) x (y + cos x) = cos x + c

Our derived solution matches option (A).

The correct option is (A) x (y + cos x) = sin x + c.

Given:

The differential equation $(e^x + 1) y \, dy = (y + 1) e^x \, dx$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is:

$(e^x + 1) y \, dy = (y + 1) e^x \, dx$

This is a separable differential equation. To separate the variables $x$ and $y$, divide both sides by $(e^x + 1)(y + 1)$, assuming $e^x + 1 \neq 0$ (which is always true for real $x$) and $y + 1 \neq 0$ (i.e., $y \neq -1$).

$\frac{(e^x + 1) y}{(e^x + 1)(y + 1)} \, dy = \frac{(y + 1) e^x}{(e^x + 1)(y + 1)} \, dx$

$\frac{y}{y + 1} \, dy = \frac{e^x}{e^x + 1} \, dx$

Now, integrate both sides of the separated equation:

$\int \frac{y}{y + 1} \, dy = \int \frac{e^x}{e^x + 1} \, dx$

Evaluate the integral on the left side $\int \frac{y}{y + 1} \, dy$.

We can rewrite the integrand as $\frac{y+1-1}{y+1} = 1 - \frac{1}{y+1}$.

$\int \left(1 - \frac{1}{y + 1}\right) \, dy = \int 1 \, dy - \int \frac{1}{y + 1} \, dy = y - \log|y + 1| + C_1$

Evaluate the integral on the right side $\int \frac{e^x}{e^x + 1} \, dx$.

Let $u = e^x + 1$. Then $du = e^x \, dx$.

The integral becomes $\int \frac{1}{u} \, du = \log|u| + C_2$.

Substitute back $u = e^x + 1$:

$\int \frac{e^x}{e^x + 1} \, dx = \log|e^x + 1| + C_2$

Since $e^x > 0$, $e^x + 1 > 1$, so $|e^x + 1| = e^x + 1$.

$\int \frac{e^x}{e^x + 1} \, dx = \log(e^x + 1) + C_2$

Combine the results of the integration:

$y - \log|y + 1| = \log(e^x + 1) + (C_2 - C_1)$

Let $C = C_2 - C_1$, which is an arbitrary constant.

$y - \log|y + 1| = \log(e^x + 1) + C$

Rearrange the terms to match the options. Move the logarithmic terms to the right side:

$y = \log|y + 1| + \log(e^x + 1) + C$

Using the property of logarithms $\log a + \log b = \log (ab)$:

$y = \log(|y + 1| (e^x + 1)) + C$

The constant of integration can be written as $\log k$. We can adjust the constant on either side. Let's check option (C): $y = \log \{k (y + 1) (e^x + 1)\}$.

If $y = \log \{k (y + 1) (e^x + 1)\}$, then exponentiating both sides gives:

$e^y = k (y + 1) (e^x + 1)$

Let's try to manipulate our derived solution $y = \log|y + 1| + \log(e^x + 1) + C$.

$y - C = \log(|y + 1| (e^x + 1))$

Exponentiating both sides:

$e^{y - C} = |y + 1| (e^x + 1)$

$e^y e^{-C} = |y + 1| (e^x + 1)$

Let $K = e^{-C}$. Since $C$ is an arbitrary constant, $K$ is an arbitrary positive constant. The absolute value can be absorbed into the constant, giving $y+1 = K (e^x+1) e^y$. This doesn't look like the options.

Let's re-examine the integration steps and the option forms.

The integration result was $y - \log|y + 1| = \log(e^x + 1) + C$.

Rewrite this as:

$y - C = \log|y + 1| + \log(e^x + 1)$

$y - C = \log(|y + 1|(e^x + 1))$

Taking the constant on the log side, say $\log K'$:

$\log|y + 1| + \log(e^x + 1) = y - C'$

$\log(|y + 1|(e^x + 1)) = y - C'$

This still leads to $e^{y-C'} = |y+1|(e^x+1)$.

Let's check the options by differentiating them. This is often easier for MCQ problems.

Consider option (C): $y = \log \{k (y + 1) (e^x + 1)\}$

Exponentiate both sides: $e^y = k (y + 1) (e^x + 1)$

Differentiate with respect to $x$ (using product rule on the right side):

$\frac{d}{dx}(e^y) = k \frac{d}{dx}((y + 1) (e^x + 1))$

$e^y \frac{dy}{dx} = k \left[ \frac{d}{dx}(y+1) (e^x+1) + (y+1) \frac{d}{dx}(e^x+1) \right]$

$e^y \frac{dy}{dx} = k \left[ \frac{dy}{dx} (e^x+1) + (y+1) e^x \right]$

$e^y \frac{dy}{dx} = k (e^x+1) \frac{dy}{dx} + k (y+1) e^x$

Rearrange terms involving $\frac{dy}{dx}$:

$e^y \frac{dy}{dx} - k (e^x+1) \frac{dy}{dx} = k (y+1) e^x$

$\frac{dy}{dx} (e^y - k(e^x+1)) = k (y+1) e^x$

$\frac{dy}{dx} = \frac{k (y+1) e^x}{e^y - k(e^x+1)}$

This does not immediately look like the original differential equation $\frac{dy}{dx} = \frac{(y + 1) e^x}{(e^x + 1) y}$. Let's substitute $k = \frac{e^y}{(y+1)(e^x+1)}$ from $e^y = k (y+1)(e^x+1)$.

$e^y - k(e^x+1) = e^y - \frac{e^y}{(y+1)(e^x+1)}(e^x+1) = e^y - \frac{e^y}{y+1} = e^y \left( 1 - \frac{1}{y+1} \right) = e^y \left( \frac{y+1-1}{y+1} \right) = e^y \frac{y}{y+1}$

So, $\frac{dy}{dx} = \frac{k (y+1) e^x}{e^y \frac{y}{y+1}} = \frac{k (y+1)^2 e^x}{e^y y}$.

This is still not the original equation $\frac{dy}{dx} = \frac{(y + 1) e^x}{(e^x + 1) y}$. There might be an issue with option (C) or its interpretation.

Let's reconsider our integrated form: $y - \log|y + 1| = \log(e^x + 1) + C$.

Rearrange: $y - C = \log|y + 1| + \log(e^x + 1)$.

Exponentiate: $e^{y-C} = |y+1|(e^x+1)$.

Let $e^{-C} = K'$, where $K'$ is a positive constant. $K' e^y = |y+1|(e^x+1)$.

This can be written as $\frac{e^y}{|y+1|} = K'(e^x+1)$.

Or $\frac{e^y}{y+1} = K(e^x+1)$ where $K$ is an arbitrary non-zero constant (absorbing the absolute value and sign).

This form is not in the options.

Let's look at the options again and our integrated result $y - \log|y + 1| = \log(e^x + 1) + C$.

Consider option (D): $y = \log \left\{ \frac{e^x + 1}{y + 1} \right\} + k$.

Rearrange: $y - k = \log \left| \frac{e^x + 1}{y + 1} \right|$ (assuming the curly braces imply absolute value for log).

Exponentiate: $e^{y-k} = \left| \frac{e^x + 1}{y + 1} \right|$

$e^y e^{-k} = \frac{|e^x + 1|}{|y + 1|}$

Let $C' = e^{-k}$. $C' e^y = \frac{e^x + 1}{|y + 1|}$ (since $e^x+1 > 0$).

$C' e^y |y+1| = e^x + 1$. This is not the original equation.

Let's go back to the integral result $y - \log|y + 1| = \log(e^x + 1) + C$.

Rearrange terms involving log to one side: $y - C = \log|y + 1| + \log(e^x + 1)$

$y - C = \log(|y + 1|(e^x + 1))$

Let's try to manipulate the answer from option (A): $(y + 1) = k (e^x + 1)$. This is a linear relationship, not containing exponentials of $y$. Differentiating this gives $\frac{dy}{dx} = k e^x$. Substituting $k = \frac{y+1}{e^x+1}$ into the derivative gives $\frac{dy}{dx} = \frac{y+1}{e^x+1} e^x = \frac{(y+1)e^x}{e^x+1}$. The original equation is $\frac{dy}{dx} = \frac{(y + 1) e^x}{(e^x + 1) y}$. This requires an extra $1/y$ factor on the right side. So option (A) is incorrect.

Let's look at option (C) again: $y = \log \{k (y + 1) (e^x + 1)\}$.

This implies $e^y = k(y+1)(e^x+1)$.

Let's rearrange our integrated result: $y - \log|y+1| - \log(e^x+1) = C$.

$y = \log|y+1| + \log(e^x+1) + C$.

Let the constant $C = \log|K|$, where $K$ is a non-zero arbitrary constant.

$y = \log|y+1| + \log(e^x+1) + \log|K|$

$y = \log(|K(y+1)(e^x+1)|)$

If we assume $y+1$ can be positive or negative, the absolute value is needed. However, often in solutions, the absolute value is omitted, implying a choice of interval. If we assume $y+1 > 0$ and $K>0$, then $|K(y+1)(e^x+1)| = K(y+1)(e^x+1)$.

So, $y = \log(K(y+1)(e^x+1))$ is a plausible form of the solution.

Comparing with option (C), $y = \log \{k (y + 1) (e^x + 1)\}$. This matches our derived form if we interpret the curly braces as not implying absolute value and $k$ as the arbitrary constant which might absorb the absolute value sign or be restricted to positive values.

Let's assume option (C) is the correct form and verify it by differentiating. We already did this differentiation and got $\frac{dy}{dx} = \frac{k (y+1)^2 e^x}{e^y y}$. For this to be equal to $\frac{(y + 1) e^x}{(e^x + 1) y}$, we would need $\frac{k (y+1)^2 e^x}{e^y y} = \frac{(y + 1) e^x}{(e^x + 1) y}$.

Assuming $e^x \neq 0$, $y \neq 0$, $y+1 \neq 0$, we can simplify:

$k (y+1) = \frac{e^y}{e^x+1}$

$k (y+1)(e^x+1) = e^y$

This is exactly the equation we differentiated ($e^y = k (y + 1) (e^x + 1)$). This confirms that $y = \log \{k (y + 1) (e^x + 1)\}$ is indeed the solution, provided the interpretation of the curly braces allows for the constant $k$ to make the argument of the logarithm positive, or if the absolute value is implicitly understood.

The correct option is (C) y = log {k (y + 1) (e^x + 1)}.

Given:

The differential equation $\frac{dy}{dx} = e^{x−y} + x^2 e^{-y}$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is:

$\frac{dy}{dx} = e^{x−y} + x^2 e^{-y}$

Using the property of exponents $e^{a-b} = e^a e^{-b}$, we can rewrite the right side:

$\frac{dy}{dx} = e^x e^{-y} + x^2 e^{-y}$

Factor out $e^{-y}$ from the terms on the right side:

$\frac{dy}{dx} = e^{-y} (e^x + x^2)$

This is a separable differential equation. To separate the variables $x$ and $y$, multiply both sides by $e^y$ and by $dx$:

$\frac{dy}{dx} \cdot e^y = e^{-y} (e^x + x^2) \cdot e^y$

$e^y \frac{dy}{dx} = e^x + x^2$

$e^y \, dy = (e^x + x^2) \, dx$

Now, integrate both sides of the separated equation:

$\int e^y \, dy = \int (e^x + x^2) \, dx$

Evaluate the integral on the left side:

$\int e^y \, dy = e^y$

Evaluate the integral on the right side:

$\int (e^x + x^2) \, dx = \int e^x \, dx + \int x^2 \, dx = e^x + \frac{x^3}{3}$

Combine the results of the integration and add the constant of integration, $c$:

$e^y = e^x + \frac{x^3}{3} + c$

This is the general solution of the given differential equation. We can rearrange it to match the options.

Move the $e^x$ term to the left side:

$e^y - e^x = \frac{x^3}{3} + c$

Comparing this general solution with the given options:

(A) y = e^x–y – x² e^–y + c (This form involves y on both sides and is not a standard explicit or implicit solution form derived.)

(B) $e^y - e^x = \frac{x^3}{3} + c$

(C) $e^x + e^y = \frac{x^3}{3} + c$

(D) $e^x - e^y = \frac{x^3}{3} + c$ (This is $-(e^y - e^x) = \frac{x^3}{3} + c$, so $e^y - e^x = -(\frac{x^3}{3} + c) = -\frac{x^3}{3} - c$. This is the same form as (B) with a different constant, $c' = -c$.)

Both options (B) and (D) represent the same family of solutions, differing only by the sign of the arbitrary constant. However, option (B) exactly matches the form we derived directly.

The correct option is (B) $e^y - e^x = \frac{x^3}{3} + c$.

Given:

The differential equation $\frac{dy}{dx} + \frac{2xy}{1 + x^2} = \frac{1}{(1 + x^2)^2}$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is:

$\frac{dy}{dx} + \frac{2x}{1 + x^2} y = \frac{1}{(1 + x^2)^2}$

This is a first-order linear differential equation in the standard form $\frac{dy}{dx} + P(x)y = Q(x)$.

Comparing, we identify $P(x)$ and $Q(x)$:

$P(x) = \frac{2x}{1 + x^2}$

$Q(x) = \frac{1}{(1 + x^2)^2}$

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

$IF = e^{\int \frac{2x}{1 + x^2} \, dx}$

Evaluate the integral $\int \frac{2x}{1 + x^2} \, dx$.

Let $u = 1 + x^2$. Then the differential $du = 2x \, dx$.

The integral becomes $\int \frac{1}{u} \, du = \log|u|$.

Substitute back $u = 1 + x^2$: $\int \frac{2x}{1 + x^2} \, dx = \log|1 + x^2|$.

Since $1 + x^2 > 0$ for all real $x$, we have $\log(1 + x^2)$.

For the integrating factor, we can take the constant of integration to be 0.

Substitute the result of the integral back into the formula for the integrating factor:

$IF = e^{\log(1 + x^2)}$

Using the property $e^{\log a} = a$ (for $a>0$):

$IF = 1 + x^2$

The general solution of a first-order linear differential equation is given by:

$y \cdot (IF) = \int Q(x) \cdot (IF) \, dx + C$

where $C$ (or $c$ in the options) is the arbitrary constant of integration.

Substitute the values of $y$, $IF$, and $Q(x)$:

$y \cdot (1 + x^2) = \int \frac{1}{(1 + x^2)^2} \cdot (1 + x^2) \, dx + c$

$y (1 + x^2) = \int \frac{1}{1 + x^2} \, dx + c$

Evaluate the integral $\int \frac{1}{1 + x^2} \, dx$:

$\int \frac{1}{1 + x^2} \, dx = \tan^{-1} x$

Substitute the result of the integral back into the general solution equation:

$y (1 + x^2) = \tan^{-1} x + c$

This is the general solution of the given differential equation.

Comparing this general solution with the given options:

(A) y (1 + x²) = c + tan^–1 x

(B) $\frac{y}{1 + x^2} = c + \tan^{-1} x$

(C) y log (1 + x²) = c + tan^–1 x

(D) y (1 + x²) = c + sin^–1 x

Our derived solution matches option (A).

The correct option is (A) y (1 + x²) = c + tan^–1 x.

(i) The given differential equation is $\frac{d^2y}{dx^2} + e^{\frac{dy}{dx}} = 0$.

The highest order derivative is $\frac{d^2y}{dx^2}$, so the order is 2.

The term $e^{\frac{dy}{dx}}$ involves a transcendental function of a derivative ($\frac{dy}{dx}$). The differential equation cannot be expressed as a polynomial in its derivatives.

Therefore, the degree of the differential equation is not defined.

(ii) The given differential equation is $\sqrt{1 + \left( \frac{dy}{dx} \right)^2} = x$.

To find the degree, we must make the equation a polynomial in its derivatives by removing the radical. Square both sides:

$\left(\sqrt{1 + \left( \frac{dy}{dx} \right)^2}\right)^2 = x^2$

$1 + \left( \frac{dy}{dx} \right)^2 = x^2$

This is a polynomial equation in the derivative $\frac{dy}{dx}$.

The highest order derivative is $\frac{dy}{dx}$, so the order is 1.

The highest power of the highest order derivative ($\frac{dy}{dx}$) is 2.

Therefore, the degree of the differential equation is 2.

(iii) The number of arbitrary constants in the general solution of a differential equation is equal to its order.

For a differential equation of order three, the number of arbitrary constants in its general solution is three.

(iv) The given differential equation is $\frac{dy}{dx} + \frac{y}{x \log x} = \frac{1}{x}$.

This equation can be written in the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = \frac{1}{x \log x}$ and $Q(x) = \frac{1}{x}$.

This is the standard form of a first-order linear differential equation.

(v) The given differential equation is $\frac{dy}{dx} + P_1x = Q_1$.

Rearrange the equation: $\frac{dy}{dx} = Q_1 - P_1x$.

This is a first-order differential equation where the right side is a function of $x$ (or constants $P_1, Q_1$).

Integrate both sides with respect to $x$:

$\int \frac{dy}{dx} \, dx = \int (Q_1 - P_1x) \, dx$

$y = \int Q_1 \, dx - \int P_1x \, dx$

$y = Q_1 x - P_1 \frac{x^2}{2} + C$

where C is the arbitrary constant of integration.

The general solution is $y = Q_1 x - \frac{P_1}{2} x^2 + C$.

(vi) The given differential equation is $x \frac{dy}{dx} + 2y = x^2$.

This is a first-order linear differential equation. Write it in standard form $\frac{dy}{dx} + P(x)y = Q(x)$ by dividing by $x$ (assuming $x \neq 0$):

$\frac{dy}{dx} + \frac{2}{x} y = x$

Here, $P(x) = \frac{2}{x}$ and $Q(x) = x$.

The integrating factor is $IF = e^{\int P(x) dx} = e^{\int \frac{2}{x} dx} = e^{2 \int \frac{1}{x} dx} = e^{2 \log|x|} = e^{\log|x|^2} = |x|^2 = x^2$. We use $IF=x^2$ (valid for $x \neq 0$).

The general solution is $y \cdot (IF) = \int Q(x) \cdot (IF) \, dx + C$:

$y \cdot x^2 = \int x \cdot x^2 \, dx + C$

$x^2 y = \int x^3 \, dx + C$

$x^2 y = \frac{x^4}{4} + C$

The solution is $y = \frac{x^4}{4x^2} + \frac{C}{x^2} = \frac{x^2}{4} + \frac{C}{x^2}$.

(vii) The given differential equation is $(1 + x^2) \frac{dy}{dx} +2xy – 4x^2 = 0$.

This is a first-order linear differential equation. Write it in standard form $\frac{dy}{dx} + P(x)y = Q(x)$ by rearranging and dividing by $(1 + x^2)$:

$(1 + x^2) \frac{dy}{dx} + 2xy = 4x^2$

$\frac{dy}{dx} + \frac{2x}{1 + x^2} y = \frac{4x^2}{1 + x^2}$

Here, $P(x) = \frac{2x}{1 + x^2}$ and $Q(x) = \frac{4x^2}{1 + x^2}$.

The integrating factor is $IF = e^{\int P(x) dx} = e^{\int \frac{2x}{1 + x^2} dx}$. Let $u = 1+x^2$, $du = 2x dx$. $\int \frac{du}{u} = \log|u| = \log(1+x^2)$.

$IF = e^{\log(1 + x^2)} = 1 + x^2$.

The general solution is $y \cdot (IF) = \int Q(x) \cdot (IF) \, dx + C$:

$y (1 + x^2) = \int \frac{4x^2}{1 + x^2} (1 + x^2) \, dx + C$

$y (1 + x^2) = \int 4x^2 \, dx + C$

$y (1 + x^2) = 4 \int x^2 \, dx + C$

$y (1 + x^2) = 4 \frac{x^3}{3} + C$

The solution is $y (1 + x^2) = \frac{4x^3}{3} + C$.

(viii) The given differential equation is $y dx + (x + xy) dy = 0$.

Factor the term with $dy$: $y dx + x(1 + y) dy = 0$.

This is a separable differential equation. Separate the variables by dividing by $xy$ (assuming $x \neq 0, y \neq 0$):

$\frac{y \, dx}{xy} + \frac{x(1 + y) \, dy}{xy} = 0$

$\frac{dx}{x} + \frac{1 + y}{y} \, dy = 0$

$\frac{dx}{x} + \left(\frac{1}{y} + 1\right) \, dy = 0$

Integrate both sides:

$\int \frac{1}{x} \, dx + \int \left(\frac{1}{y} + 1\right) \, dy = \int 0 \, dx$

$\log|x| + \log|y| + y = C$

Using logarithm properties: $\log|xy| + y = C$.

The solution is $\log|xy| + y = C$.

(ix) The given differential equation is $\frac{dy}{dx} + y = \sin x$.

This is a first-order linear differential equation in standard form $\frac{dy}{dx} + P(x)y = Q(x)$, with $P(x) = 1$ and $Q(x) = \sin x$.

The integrating factor is $IF = e^{\int 1 \, dx} = e^x$.

The general solution is $y \cdot (IF) = \int Q(x) \cdot (IF) \, dx + C$:

$y e^x = \int \sin x \cdot e^x \, dx + C$

Evaluate the integral $\int e^x \sin x \, dx$. Use integration by parts twice or a formula. The formula is $\int e^{ax} \sin(bx) \, dx = \frac{e^{ax}}{a^2 + b^2} (a \sin(bx) - b \cos(bx))$. Here $a=1, b=1$.

$\int e^x \sin x \, dx = \frac{e^x}{1^2 + 1^2} (1 \sin x - 1 \cos x) = \frac{e^x}{2} (\sin x - \cos x)$.

So,

$y e^x = \frac{e^x}{2} (\sin x - \cos x) + C$

Divide by $e^x$:

$y = \frac{1}{2} (\sin x - \cos x) + C e^{-x}$.

The general solution is $y = \frac{1}{2}(\sin x - \cos x) + C e^{-x}$.

(x) The given differential equation is $\cot y \, dx = x \, dy$.

This is a separable differential equation. Separate the variables by dividing by $x \cot y$ (assuming $x \neq 0, \cot y \neq 0$):

$\frac{\cot y \, dx}{x \cot y} = \frac{x \, dy}{x \cot y}$

$\frac{dx}{x} = \frac{dy}{\cot y}$

$\frac{dx}{x} = \tan y \, dy$

Integrate both sides:

$\int \frac{1}{x} \, dx = \int \tan y \, dy$

$\log|x| = \log|\sec y| + C$

Rearrange: $\log|x| - \log|\sec y| = C$.

$\log\left|\frac{x}{\sec y}\right| = C$

Exponentiate: $\left|\frac{x}{\sec y}\right| = e^C$. Let $k = \pm e^C$.

$\frac{x}{\sec y} = k$

$x = k \sec y$. This can also be written as $x = k \frac{1}{\cos y}$, or $x \cos y = k$.

The solution is $x \cos y = k$.

(xi) The given differential equation is $\frac{dy}{dx} + y = \frac{1 + y}{x}$.

Rewrite the equation in standard form $\frac{dy}{dx} + P(x)y = Q(x)$:

$\frac{dy}{dx} + y = \frac{1}{x} + \frac{y}{x}$

$\frac{dy}{dx} + y - \frac{y}{x} = \frac{1}{x}$

$\frac{dy}{dx} + y \left(1 - \frac{1}{x}\right) = \frac{1}{x}$

$\frac{dy}{dx} + \left(\frac{x-1}{x}\right) y = \frac{1}{x}$

Here, $P(x) = \frac{x-1}{x} = 1 - \frac{1}{x}$.

The integrating factor is $IF = e^{\int P(x) dx} = e^{\int (1 - \frac{1}{x}) dx}$.

$\int (1 - \frac{1}{x}) dx = x - \log|x|$.

$IF = e^{x - \log|x|} = e^x \cdot e^{-\log|x|} = e^x \cdot e^{\log|x|^{-1}} = e^x \cdot \frac{1}{|x|}$.

Assuming $x > 0$, the integrating factor is $e^x \cdot \frac{1}{x} = \frac{e^x}{x}$.

The integrating factor is $\frac{e^x}{x}$.

(i) The given differential equation is $\frac{d^2y}{dx^2} + e^{\frac{dy}{dx}} = 0$. This equation involves a transcendental function of a derivative, so its degree is not defined. The integrating factor formula $e^{\int p_1 dy}$ is associated with a first-order linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$, which is different from the given form $\frac{dy}{dx} + p_1x = Q_1$. The statement connects an incorrect IF formula to the given DE form.

The statement is False.

(ii) The differential equation is $\sqrt{1 + \left( \frac{dy}{dx} \right)^2} = x$. Squaring both sides gives $1 + \left( \frac{dy}{dx} \right)^2 = x^2$. This is a polynomial in the derivative $\frac{dy}{dx}$. The highest order derivative is $\frac{dy}{dx}$ (order 1), and its highest power is 2. Thus, the degree is 2. Interpreting the blank as claiming the degree is 2, the statement is true.

The statement is True.

(iii) The number of arbitrary constants in the general solution of a differential equation is equal to its order. For an order three differential equation, the number of arbitrary constants is three. Interpreting the blank as claiming the number is three, the statement is true.

The statement is True.

(iv) The equation $\frac{dy}{dx} + \frac{y}{x \log x} = \frac{1}{x}$ is of the form $\frac{dy}{dx} + P(x)y = Q(x)$, which is the standard form of a first-order linear differential equation. Interpreting the blank as claiming it is a linear type, the statement is true.

The statement is True.

(v) The general solution of a linear differential equation $\frac{dy}{dx} + P(x)y = Q(x)$ is $y \cdot IF = \int Q(x) \cdot IF \, dx + C$. If the equation were $\frac{dx}{dy} + P_1(y)x = Q_1(y)$, the solution would be $x \cdot IF = \int Q_1(y) \cdot IF \, dy + C$. The statement "x.I.F = (I.F) × Q₁ dy" is not the correct general solution formula; it is missing the integral and the constant of integration, and the equation form $\frac{dy}{dx} + P_1x = Q_1$ is not a standard linear form where $x \cdot IF$ is used.

The statement is False.

(vi) The equation of the family of circles is $x^2 + (y – a)^2 = a^2$, which simplifies to $x^2 + y^2 - 2ay = 0$. This equation contains one arbitrary constant ($a$). The order of the differential equation representing a family of curves is equal to the number of essential arbitrary constants. Thus, the order is 1. The statement claims the order is two.

The statement is False.

(vii) The differential equation $\frac{dy}{dx} = \left( \frac{y}{x} \right)^{\frac{1}{3}}$ is homogeneous of degree 0. Substituting $y=vx$ and solving gives the solution $y^{2/3} - x^{2/3} = c$ (or $x^{2/3} - y^{2/3} = k$). As shown in the analysis for Question 76 (vii), the solution is $x^{2/3} - y^{2/3} = c'$, which is equivalent to $y^{2/3} - x^{2/3} = -c'$. Since $c'$ is arbitrary, $-c'$ is also arbitrary, so $y^{2/3} - x^{2/3} = c$ is a valid form of the general solution.

The statement is True.

(viii) As shown in the solution for Question 37, the differential equation for which $y = e^x (A \cos x + B \sin x)$ is a solution is $\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0$. The statement claims the DE is $\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0$. (Wait, the DE in Q37 was $\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 2y = 0$. Let me recheck the derivative in Q37. Ah, $y=e^{-x}(A \cos x + B \sin x)$ has the DE $\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 2y = 0$. This question has $y = e^x (A \cos x + B \sin x)$. Let's derive it quickly: $y' = e^x(A \cos x + B \sin x) + e^x(-A \sin x + B \cos x) = y + e^x(-A \sin x + B \cos x)$. $y'' = y' + e^x(-A \sin x + B \cos x) + e^x(-A \cos x - B \sin x) = y' + (y'-y) - y = 2y' - 2y$. So $\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0$. This matches the statement.)

The statement is True.

(ix) The differential equation $\frac{dy}{dx} = \frac{x + 2y}{x} = 1 + 2\frac{y}{x}$ is homogeneous of degree 0. Substituting $y=vx$ and solving gives the solution $x + y = kx^2$, as shown in the analysis for Question 76 (ix).

The statement is True.

(x) The differential equation $\frac{x\;dy}{dx} = y + x \tan \frac{y}{x}$ can be written as $\frac{dy}{dx} = \frac{y}{x} + \tan \frac{y}{x}$, which is homogeneous of degree 0. Substituting $y=vx$ and solving gives the solution $\sin \left( \frac{y}{x} \right) = cx$, as shown in the analysis for Question 76 (x).

The statement is True.

(xi) Non-horizontal lines in a plane are all lines except those with zero slope ($y=constant$). This set includes slanted lines ($y=mx+c$, $m\neq 0$) and vertical lines ($x=k$). The differential equation $\frac{d^2x}{dy^2} = 0$ has the general solution $x = m'y + c'$. This family of curves includes vertical lines (when $m'=0$, $x=c'$) and slanted lines (when $m' \neq 0$, $x=m'y+c'$). This family precisely represents all non-horizontal lines.

The statement is True.