Applied Mathematics for Class 11th & 12th (Concepts and Questions) | ||
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12th | Concepts | Questions |
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Differential Equations | Formulating and Solving Differential Equations | Application of Differential Equations |
Chapter 5 Differential Equations and Modeling (Concepts)
Welcome to this crucial chapter introducing Differential Equations (DEs), a powerful mathematical framework for describing and analyzing systems that change over time or space. Unlike algebraic equations which solve for specific numerical values, differential equations involve relationships between functions and their derivatives (rates of change). This makes them indispensable tools in virtually every quantitative field, allowing us to model phenomena ranging from population dynamics and chemical reactions to electrical circuits and economic forecasting. This chapter aims to equip you with the fundamental vocabulary, classification methods, and core techniques for solving basic types of differential equations commonly encountered in Applied Mathematics, emphasizing their role as instruments for modeling real-world processes.
We begin by formally defining a differential equation as any equation containing derivatives of one or more dependent variables with respect to one or more independent variables. We will learn to classify DEs based on their order (the order of the highest derivative appearing in the equation) and degree (the power of the highest-order derivative after the equation has been cleared of radicals and fractional powers involving derivatives). While the broader world includes Partial Differential Equations (PDEs) involving derivatives with respect to multiple independent variables, our focus in this introductory context will be solely on Ordinary Differential Equations (ODEs), which involve derivatives with respect to only a single independent variable. A key concept is understanding the nature of solutions: a general solution represents a family of functions satisfying the DE and typically contains arbitrary constants (the number of which equals the order of the DE), while a particular solution is obtained by using specific initial or boundary conditions to determine the values of these constants, tailoring the solution to a specific scenario.
Conceptually, we understand that a differential equation often arises from eliminating arbitrary constants from a relation representing a family of curves. However, our primary practical focus will be on techniques for solving given first-order, first-degree differential equations. Two fundamental methods applicable to a wide range of problems in applied contexts are:
- Variables Separable Method: This technique applies when the differential equation can be algebraically rearranged so that all terms involving the dependent variable (say $y$) and its differential $dy$ are on one side of the equation, and all terms involving the independent variable (say $x$) and $dx$ are on the other side. This leads to an equation of the form $g(y)dy = f(x)dx$, which can then be solved by integrating both sides: $\int g(y)dy = \int f(x)dx + C$.
- Linear Differential Equations: This method addresses equations of the specific form $\frac{dy}{dx} + P(x)y = Q(x)$. The solution involves calculating an Integrating Factor (IF), defined as $IF = e^{\int P(x)dx}$. Multiplying the entire DE by the IF makes the left side the derivative of a product, leading to the general solution: $y \times (IF) = \int (Q(x) \times IF) dx + C$. Successfully applying this method often relies on correctly evaluating the integrals involved in finding the IF and the final solution.
Beyond the mechanics of solving, the true power of DEs lies in mathematical modeling. Differential equations emerge naturally when we describe processes based on their rates of change. A cornerstone application explored is modeling Exponential Growth and Decay. Phenomena like unrestricted population growth, compound interest compounded continuously, or radioactive decay are often described by the differential equation $\frac{dN}{dt} = kN$, where $N$ is the quantity, $t$ is time, and $k$ is the constant of proportionality (positive for growth, negative for decay). Solving this separable DE leads to the exponential function $N(t) = N_0 e^{kt}$. Other potential application areas, such as simple models based on Newton's Law of Cooling or basic learning curves, might also be introduced conceptually. The overarching skill emphasized is the ability to translate a word problem describing a rate of change into an appropriate differential equation, solve it using the methods learned, and then apply any given initial conditions to find the specific particular solution that models the real-world situation under investigation.
Differential Equations
A differential equation is an equation that relates one or more unknown functions and their derivatives with respect to one or more independent variables. Essentially, it's an equation involving derivatives. Differential equations are fundamental mathematical tools used to describe how physical, biological, economic, or other systems change over time or space. They capture the dynamic relationships between quantities and their rates of change.
Types of Differential Equations
Differential equations are broadly classified based on the type of derivatives involved:
Ordinary Differential Equations (ODEs)
An ordinary differential equation (ODE) involves only one independent variable and one or more derivatives with respect to that single independent variable.
For example, if $y$ is a function of $x$, an ODE might involve $\frac{dy}{dx}$, $\frac{d^2y}{dx^2}$, etc.
Examples:
- $\frac{dy}{dx} = 2x$
- $\frac{d^2y}{dx^2} + \frac{dy}{dx} + y = 0$
- $(\frac{dy}{dt})^2 + ty = \sin t$
Partial Differential Equations (PDEs)
A partial differential equation (PDE) involves two or more independent variables and partial derivatives with respect to those variables.
For example, if $u$ is a function of $x$ and $t$, a PDE might involve $\frac{\partial u}{\partial x}$, $\frac{\partial u}{\partial t}$, $\frac{\partial^2 u}{\partial x^2}$, $\frac{\partial^2 u}{\partial t^2}$, $\frac{\partial^2 u}{\partial x \partial t}$, etc.
Examples:
- Heat equation: $\frac{\partial u}{\partial t} = c^2 \frac{\partial^2 u}{\partial x^2}$
- Laplace's equation: $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$
- Wave equation: $\frac{\partial^2 u}{\partial t^2} = c^2 (\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2})$
Note: In this chapter, we will primarily focus on Ordinary Differential Equations (ODEs) and their applications.
Order of a Differential Equation
The order of a differential equation is defined as the order of the highest derivative that appears in the equation. It indicates the complexity of the rate of change described by the equation.
Example 1: Consider the equation $\frac{dy}{dx} = \sin x$.
The highest derivative is $\frac{dy}{dx}$, which is a first derivative. Therefore, the order of this differential equation is 1.
Example 2: Consider the equation $\frac{d^2y}{dx^2} + 5 \frac{dy}{dx} + 6y = 0$.
The derivatives present are $\frac{d^2y}{dx^2}$ (second derivative) and $\frac{dy}{dx}$ (first derivative). The highest order is 2 (from $\frac{d^2y}{dx^2}$). Therefore, the order of this differential equation is 2.
Example 3: Consider the equation $(\frac{d^3y}{dx^3})^2 + (\frac{dy}{dx})^4 + y = x$.
The derivatives present are $\frac{d^3y}{dx^3}$ (third derivative) and $\frac{dy}{dx}$ (first derivative). The highest order is 3 (from $\frac{d^3y}{dx^3}$). Therefore, the order of this differential equation is 3. The powers of the derivatives do not affect the order.
Degree of a Differential Equation
The degree of a differential equation is defined as the power of the highest order derivative appearing in the equation, provided that the equation can be expressed as a polynomial in the derivatives.
To find the degree, the differential equation must first be cleared of any radicals (like square roots) or fractions (where derivatives are in the denominator) with respect to the derivatives. If the equation cannot be expressed as a polynomial in its derivatives (e.g., if a derivative is inside a sine or exponential function like $\sin(\frac{dy}{dx})$ or $e^{\frac{d^2y}{dx^2}}$), then the degree is not defined.
Example 1: Consider the equation $(\frac{dy}{dx})^2 + y = x$.
The highest order derivative is $\frac{dy}{dx}$ (order 1). The equation is already a polynomial in the derivatives. The power of the highest order derivative $(\frac{dy}{dx})$ is 2. Thus, the degree is 2.
Example 2: Consider the equation $\frac{d^2y}{dx^2} + \frac{dy}{dx} + y = 0$.
The highest order derivative is $\frac{d^2y}{dx^2}$ (order 2). The equation is a polynomial in the derivatives. The power of the highest order derivative $(\frac{d^2y}{dx^2})$ is 1. Thus, the degree is 1.
Example 3: Consider the equation $\frac{d^2y}{dx^2} = \sqrt{1 + (\frac{dy}{dx})^2}$.
This equation involves a radical containing derivatives. To find the degree, we first need to eliminate the radical by squaring both sides:
$(\frac{d^2y}{dx^2})^2 = 1 + (\frac{dy}{dx})^2$
... (1)
Now the equation is free from radicals and is a polynomial in the derivatives. The highest order derivative is $\frac{d^2y}{dx^2}$ (order 2). The power of this highest order derivative is 2. Thus, the degree is 2.
Example 4: Consider the equation $\sin(\frac{dy}{dx}) = x$.
The highest order derivative is $\frac{dy}{dx}$. However, this derivative is inside a trigonometric function. This equation cannot be written as a polynomial in derivatives. Therefore, the degree of this differential equation is not defined.
Solution of a Differential Equation
A solution of a differential equation is a function that, when substituted into the differential equation along with its derivatives, satisfies the equation identically. In simpler terms, a solution is a function $y = f(x)$ (or $u = f(x,t)$ for PDEs) that makes the differential equation true for all values of the independent variable(s) in a given domain.
There are typically two types of solutions we discuss:
General Solution
A general solution of a differential equation is a solution that contains one or more arbitrary constants. For an ordinary differential equation of order $n$, the general solution typically contains $n$ arbitrary independent constants. This solution represents a family of curves, each corresponding to a specific value of the arbitrary constant(s).
Particular Solution
A particular solution is obtained from the general solution by assigning specific values to the arbitrary constants. These specific values are usually determined by using additional conditions, such as initial conditions (conditions specified at a single point, often related to time, e.g., $y(t_0) = y_0$ and $y'(t_0) = y_0'$) or boundary conditions (conditions specified at different points in space or time).
A singular solution is a solution that cannot be obtained from the general solution by assigning specific values to the arbitrary constants. (This is less common in introductory applied maths).
Examples of Solutions
Example 1. Show that $y = Ce^{2x}$ is a solution of the differential equation $\frac{dy}{dx} - 2y = 0$, where C is an arbitrary constant.
Answer:
Given the function $y = Ce^{2x}$. This function represents a family of curves, parameterized by the constant $C$. We need to verify if this function satisfies the given differential equation $\frac{dy}{dx} - 2y = 0$.
First, let's find the derivative of $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(Ce^{2x})$
$\frac{dy}{dx} = C \frac{d}{dx}(e^{2x})$
$\frac{dy}{dx} = C (e^{2x} \cdot \frac{d}{dx}(2x))$
$\frac{dy}{dx} = C (e^{2x} \cdot 2)$
$\frac{dy}{dx} = 2Ce^{2x}$
... (1)
Now, substitute the expression for $\frac{dy}{dx}$ from (1) and the expression for $y$ into the given differential equation $\frac{dy}{dx} - 2y = 0$:
$(2Ce^{2x}) - 2(Ce^{2x})$
$2Ce^{2x} - 2Ce^{2x}$
$0$
The left-hand side of the differential equation simplifies to 0, which is equal to the right-hand side (0).
Since the equation $\frac{dy}{dx} - 2y = 0$ is satisfied for $y = Ce^{2x}$ for any arbitrary constant $C$, $y = Ce^{2x}$ is indeed a general solution to the given differential equation. This represents a family of exponential curves.
Example 2. Verify that $y = x^2 + 3x$ is a solution to the differential equation $x\frac{dy}{dx} - (x+1)y = x^2 - 3x$.
Answer:
Given the function $y = x^2 + 3x$. We need to verify if this function satisfies the differential equation $x\frac{dy}{dx} - (x+1)y = x^2 - 3x$.
First, let's find the derivative of $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(x^2 + 3x)$
$\frac{dy}{dx} = 2x + 3$
... (1)
Now, substitute the expression for $\frac{dy}{dx}$ from (1) and the expression for $y$ into the left-hand side (LHS) of the given differential equation:
LHS = $x\frac{dy}{dx} - (x+1)y$
LHS = $x(2x+3) - (x+1)(x^2 + 3x)$
LHS = $(2x^2 + 3x) - (x(x^2 + 3x) + 1(x^2 + 3x))$
LHS = $2x^2 + 3x - (x^3 + 3x^2 + x^2 + 3x)$
LHS = $2x^2 + 3x - (x^3 + 4x^2 + 3x)$
LHS = $2x^2 + 3x - x^3 - 4x^2 - 3x$
LHS = $-x^3 + (2x^2 - 4x^2) + (3x - 3x)$
LHS = $-x^3 - 2x^2$
... (2)
Now compare the LHS (from (2)) with the right-hand side (RHS) of the given differential equation:
RHS = $x^2 - 3x$.
Comparing LHS ($-x^3 - 2x^2$) and RHS ($x^2 - 3x$), we see that LHS $\neq$ RHS.
Correction: There must be a typo in the question as provided in the input. Let's assume the differential equation was meant to be something that $y=x^2+3x$ satisfies. For instance, if the equation was $\frac{dy}{dx} - (2x+3) = 0$.
Example 2 (Corrected). Verify that $y = x^2 + 3x$ is a solution to the differential equation $\frac{dy}{dx} = 2x + 3$.
Answer:
Given the function $y = x^2 + 3x$. We need to verify if this function satisfies the differential equation $\frac{dy}{dx} = 2x + 3$.
Find the derivative of $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(x^2 + 3x)$
$\frac{dy}{dx} = 2x + 3$
... (1)
The expression for $\frac{dy}{dx}$ from (1) is $2x+3$.
The given differential equation is $\frac{dy}{dx} = 2x + 3$.
Substituting the calculated $\frac{dy}{dx}$ into the differential equation, we get:
$2x + 3 = 2x + 3$
Since the equation is satisfied, $y = x^2 + 3x$ is a solution to the differential equation $\frac{dy}{dx} = 2x + 3$. This is a particular solution as it contains no arbitrary constant.
Formulating and Solving Differential Equations
The process of applying differential equations to real-world situations involves two main steps: formulation and solving. We first translate the description of how quantities change into a differential equation (formulation), and then we find the function(s) that satisfy this equation (solving).
Formulation of Differential Equations from Word Problems
Many phenomena in nature, science, engineering, and economics can be described by mathematical models involving rates of change. Differential equations provide a powerful language for expressing these rates. Problems related to population growth or decay, radioactive decay, Newton's Law of Cooling, motion of objects, chemical reaction rates, compound interest, etc., are often modeled using differential equations.
The general steps involved in formulating a differential equation from a description are:
- Identify the Variables: Determine the dependent variable(s) (the quantity that is changing) and the independent variable(s) (the variable(s) with respect to which the rate of change is described, usually time or space).
- Express Rates of Change as Derivatives: Translate any statement about the rate of change of a quantity into a derivative. For example, "rate of change of $y$ with respect to $x$" is $\frac{dy}{dx}$.
- Write the Equation: Use the information given in the problem to establish a relationship between the variables and their derivatives, forming the differential equation.
Example: Consider the statement: "The rate of growth of a bacterial population is directly proportional to the population size at any given time."
Let $P(t)$ be the population size at time $t$. The independent variable is time, $t$. The dependent variable is the population size, $P$. The rate of growth of the population is $\frac{dP}{dt}$. The statement "directly proportional to the population size" means $\frac{dP}{dt} \propto P(t)$. Introducing a constant of proportionality, say $k$ (where $k > 0$ for growth), we can write the relationship as an equation:
$\frac{dP}{dt} = kP$
... (1)
This is a first-order ordinary differential equation that models the population growth.
Solving First Order First Degree Ordinary Differential Equations
A first-order, first-degree ordinary differential equation is one where the highest derivative is of order 1, and its power is also 1. Such equations can often be written in the form $\frac{dy}{dx} = f(x, y)$. We will discuss two common methods for solving such equations that are covered in the curriculum.
Method of Separation of Variables
A first-order differential equation $\frac{dy}{dx} = f(x, y)$ is called separable if the function $f(x, y)$ can be factored into a product of a function of $x$ only, $g(x)$, and a function of $y$ only, $h(y)$. That is, if the equation can be written in the form:
$\frac{dy}{dx} = g(x) \cdot h(y)$
... (2)
The method of separation of variables involves rearranging the equation so that all terms involving the dependent variable $y$ and $dy$ are on one side, and all terms involving the independent variable $x$ and $dx$ are on the other side. Assuming $h(y) \neq 0$, we can divide both sides by $h(y)$ and multiply by $dx$:
$\frac{1}{h(y)} dy = g(x) dx$
Once the variables are separated, we integrate both sides of the equation:
$\int \frac{1}{h(y)} dy = \int g(x) dx + C$
... (3)
where $C$ is the constant of integration. Evaluating these integrals gives the general solution to the differential equation. The key is to be able to perform the integration on both sides.
Linear Differential Equations of First Order
A first-order ordinary differential equation is said to be linear if it can be written in the standard form:
$\frac{dy}{dx} + P(x) y = Q(x)$
... (4)
where $P(x)$ and $Q(x)$ are continuous functions of the independent variable $x$ (or constants). The key features of a linear first-order equation are:
- The dependent variable $y$ and its derivative $\frac{dy}{dx}$ appear only in the first degree.
- There are no products of $y$ and $\frac{dy}{dx}$ (e.g., $y \frac{dy}{dx}$).
- $y$ or $\frac{dy}{dx}$ are not arguments of non-linear functions (e.g., $\sin y$, $e^{dy/dx}$).
To solve a linear differential equation of the form (4), we use a special function called the Integrating Factor (IF). The integrating factor is a function that, when multiplied by the differential equation, makes the left-hand side the derivative of a product.
The integrating factor (IF) for the equation $\frac{dy}{dx} + P(x) y = Q(x)$ is given by:
$\mathbf{IF = e^{\int P(x) dx}}$
... (5)
Note that we typically don't include the constant of integration when calculating the IF, as it cancels out later.
Once the integrating factor is found, we multiply both sides of the standard linear equation (4) by the IF:
$e^{\int P(x) dx} \frac{dy}{dx} + P(x) y e^{\int P(x) dx} = Q(x) e^{\int P(x) dx}$
Observe the left-hand side (LHS). It has the form $u'v + uv'$, where $u = y$ and $v = e^{\int P(x) dx}$. The derivative of $v = e^{\int P(x) dx}$ with respect to $x$ is $\frac{dv}{dx} = e^{\int P(x) dx} \cdot \frac{d}{dx}\left(\int P(x) dx\right) = e^{\int P(x) dx} \cdot P(x)$. So, the LHS is precisely the derivative of the product $y \cdot e^{\int P(x) dx}$:
$\frac{d}{dx} \left( y \cdot e^{\int P(x) dx} \right) = Q(x) e^{\int P(x) dx}$
[Using Product Rule: $(uv)' = u'v + uv'$]
Now, to solve for $y$, we integrate both sides with respect to $x$:
$\int \frac{d}{dx} \left( y \cdot e^{\int P(x) dx} \right) dx = \int Q(x) e^{\int P(x) dx} dx$
The integral of a derivative of a function is the function itself (plus a constant of integration).
$y \cdot e^{\int P(x) dx} = \int Q(x) e^{\int P(x) dx} dx + C$
[Integration reverses differentiation; add constant on RHS]
Substituting $IF = e^{\int P(x) dx}$, we get the general solution formula for a linear first-order differential equation:
$\mathbf{y \cdot (IF) = \int Q(x) \cdot (IF) dx + C}$
... (6)
This formula allows us to directly find the general solution after calculating the integrating factor and the integral on the right side.
Examples
Example 1. Solve the differential equation $\frac{dy}{dx} = \frac{x^2}{y^3}$.
Answer:
Given the differential equation:
$\frac{dy}{dx} = \frac{x^2}{y^3}$
This equation can be written as $\frac{dy}{dx} = x^2 \cdot \frac{1}{y^3}$. The right-hand side is a product of a function of $x$ only ($g(x) = x^2$) and a function of $y$ only ($h(y) = \frac{1}{y^3}$). Thus, it is a separable differential equation.
We separate the variables by moving all $y$ terms (and $dy$) to one side and all $x$ terms (and $dx$) to the other side:
$y^3 \, dy = x^2 \, dx$
... (1)
Now, integrate both sides of equation (1):
$\int y^3 \, dy = \int x^2 \, dx$
Performing the integration:
$\frac{y^{3+1}}{3+1} = \frac{x^{2+1}}{2+1} + C$
[Using $\int z^n dz = \frac{z^{n+1}}{n+1}$ and adding constant $C$]
$\frac{y^4}{4} = \frac{x^3}{3} + C$
... (2)
This is the general solution to the differential equation. We can leave it in this implicit form or rearrange it, for example, by multiplying by 12 to clear the denominators:
$12 \cdot \frac{y^4}{4} = 12 \cdot \frac{x^3}{3} + 12C$
$3y^4 = 4x^3 + 12C$
Let $K = 12C$ be a new arbitrary constant.
$3y^4 - 4x^3 = K$
... (3)
Both equation (2) and (3) represent the general solution.
Example 2. Solve the differential equation $x\frac{dy}{dx} + y = x^2$.
Answer:
Given the differential equation:
$x\frac{dy}{dx} + y = x^2$
This equation is not in the standard form of a linear differential equation $\frac{dy}{dx} + P(x) y = Q(x)$. To get it into the standard form, we divide the entire equation by $x$ (assuming $x \neq 0$):
$\frac{1}{x} \left(x\frac{dy}{dx} + y\right) = \frac{x^2}{x}$
$\frac{dy}{dx} + \frac{1}{x} y = x$
... (1)
Equation (1) is now in the standard linear form $\frac{dy}{dx} + P(x) y = Q(x)$, where $P(x) = \frac{1}{x}$ and $Q(x) = x$.
Step 1: Find the integrating factor (IF).
$IF = e^{\int P(x) dx}$
[Formula for IF]
$IF = e^{\int \frac{1}{x} dx}$
The integral of $\frac{1}{x}$ is $\log|x|$. For simplicity, assuming $x > 0$, we take $\int \frac{1}{x} dx = \log x$.
$IF = e^{\log x}$
$IF = x$
[Using property $e^{\log u} = u$]
The integrating factor is $x$.
Step 2: Use the general solution formula $y \cdot (IF) = \int Q(x) \cdot (IF) \, dx + C$.
$y \cdot (x) = \int (x) \cdot (x) \, dx + C$
$xy = \int x^2 \, dx + C$
Perform the integration on the right side:
$xy = \frac{x^{2+1}}{2+1} + C$
[Using $\int x^n dx = \frac{x^{n+1}}{n+1}$]
$xy = \frac{x^3}{3} + C$
... (2)
This is the general solution in implicit form. To express $y$ explicitly as a function of $x$, divide by $x$:
$y = \frac{1}{x} \left(\frac{x^3}{3} + C\right)$
$y = \frac{x^3}{3x} + \frac{C}{x}$
$y = \frac{x^2}{3} + \frac{C}{x}$
... (3)
The general solution is $y = \frac{x^2}{3} + \frac{C}{x}$. This solution is valid for $x \neq 0$.
Application of Differential Equations
Differential equations are indispensable tools in various fields because they provide a mathematical framework to describe and analyze dynamic processes—systems that change over time or space. By translating real-world phenomena into differential equations and finding their solutions, we can gain insights into how these systems behave, predict their future states, and understand the underlying mechanisms driving the change.
In this section, we will explore some common applications of first-order differential equations in modeling real-world scenarios.
Population Growth and Decay Models
One of the simplest yet widely applicable models involves the rate of change of a quantity being directly proportional to its current size. This is particularly useful for modeling population dynamics (growth of populations like bacteria, animals, or humans under ideal conditions) or radioactive decay.
If $P(t)$ represents the population size at time $t$, and the rate of change of the population, $\frac{dP}{dt}$, is proportional to the population size $P$, we can formulate the differential equation:
$\frac{dP}{dt} \propto P$
Introducing a constant of proportionality, $k$:
$\frac{dP}{dt} = kP$
... (1)
This is a first-order separable differential equation. We can solve it using the method of separation of variables:
Assuming $P \neq 0$, divide by $P$ and multiply by $dt$:
$\frac{dP}{P} = k \, dt$
Integrate both sides:
$\int \frac{dP}{P} = \int k \, dt$
Evaluating the integrals:
$\log|P| = kt + C'$
[Using $\int \frac{1}{u} du = \log|u|$ and adding constant $C'$]
Exponentiate both sides using base $e$:
$|P| = e^{kt + C'} = e^{kt} \cdot e^{C'}$
Remove the absolute value. The sign of $P$ is usually positive for population. Let $A = \pm e^{C'}$. Note that $P=0$ is also a solution to the original DE, which is included if $A=0$.
$P(t) = A e^{kt}$
... (2)
This is the general solution for the population model. The constant $A$ is determined by the initial population size. If $P(0) = P_0$ is the population at time $t=0$, then substituting into (2):
$P_0 = A e^{k \times 0} = A e^0 = A \times 1 = A$
So, the particular solution satisfying the initial condition $P(0) = P_0$ is:
$\mathbf{P(t) = P_0 e^{kt}}$
... (3)
- If $k > 0$, the population grows exponentially.
- If $k < 0$, the population decays exponentially (e.g., radioactive decay).
$P_0$ represents the initial amount of the quantity, and $k$ is the growth or decay constant.
Continuous Compound Interest
When interest is compounded continuously, the rate at which the amount of money in an account grows is proportional to the current amount in the account.
Let $A(t)$ be the amount of money at time $t$ (in years). Let $r$ be the annual interest rate (expressed as a decimal). The rate of change of the amount is $\frac{dA}{dt}$. According to the principle of continuous compounding, this rate is proportional to the current amount $A(t)$.
$\frac{dA}{dt} \propto A$
The constant of proportionality here is the interest rate $r$. So, the differential equation is:
$\frac{dA}{dt} = rA$
... (4)
This differential equation is identical in form to the population growth model (Equation 1), with $P$ replaced by $A$ and $k$ replaced by $r$. The solution process is the same.
Using separation of variables and integration, we arrive at the general solution:
$A(t) = A_0 e^{rt}$
... (5)
where $A_0 = A(0)$ is the initial principal amount invested at time $t=0$.
This formula gives the amount accumulated at time $t$ when a principal amount $A_0$ is invested at a continuous compound interest rate $r$.
Newton's Law of Cooling/Warming
Newton's Law of Cooling (which also applies to warming) describes how the temperature of an object changes as it exchanges heat with its surroundings. The law states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the temperature of the surrounding medium (ambient temperature), assuming the ambient temperature remains constant.
Let $T(t)$ be the temperature of the object at time $t$. Let $T_m$ be the constant ambient temperature of the surroundings. The rate of change of the object's temperature is $\frac{dT}{dt}$. The difference between the object's temperature and the ambient temperature is $(T - T_m)$.
According to Newton's Law:
$\frac{dT}{dt} \propto (T - T_m)$
Introducing a constant of proportionality, $k$:
$\frac{dT}{dt} = k(T - T_m)$
... (6)
The constant $k$ is related to the thermal properties of the object and its surroundings.
- If the object is hotter than the surroundings ($T > T_m$), then $T - T_m > 0$. For cooling to occur ($\frac{dT}{dt} < 0$), $k$ must be negative.
- If the object is colder than the surroundings ($T < T_m$), then $T - T_m < 0$. For warming to occur ($\frac{dT}{dt} > 0$), $k(T - T_m)$ must be positive. Since $T-T_m$ is negative, $k$ must again be negative.
Equation (6) is a first-order separable differential equation. Assuming $T \neq T_m$, we can separate the variables:
$\frac{dT}{T - T_m} = k \, dt$
Integrate both sides:
$\int \frac{dT}{T - T_m} = \int k \, dt$
Evaluating the integrals:
$\log|T - T_m| = kt + C'$
[Let $u=T-T_m$, $du=dT$; $\int \frac{1}{u} du = \log|u|$]
Exponentiate both sides:
$|T - T_m| = e^{kt + C'} = e^{kt} \cdot e^{C'}$
Let $A = \pm e^{C'}$. The constant $A$ will be positive if the object starts hotter than the surroundings ($T_0 > T_m$) and cools, and negative if it starts colder ($T_0 < T_m$) and warms, provided $T_m$ is not crossed. If $T=T_m$, then $\frac{dT}{dt}=0$, which is a valid constant solution included if $A=0$.
$T - T_m = A e^{kt}$
Rearranging to solve for $T(t)$:
$\mathbf{T(t) = T_m + A e^{kt}}$
... (7)
This is the general solution for Newton's Law of Cooling/Warming. The constant $A$ is determined by the initial temperature of the object. If $T(0) = T_0$ is the initial temperature at time $t=0$, then:
$T_0 = T_m + A e^{k \times 0} = T_m + A e^0 = T_m + A$
So, $A = T_0 - T_m$. Substituting this back into (7), we get the particular solution:
$\mathbf{T(t) = T_m + (T_0 - T_m) e^{kt}}$
... (8)
This form is particularly useful for solving problems involving specific initial conditions.
Other Applications
Differential equations, especially first-order ones, find applications in numerous other areas:
- Mixing Problems: Modeling the concentration of a substance (like salt or pollutant) in a tank where liquids are flowing in and out. The rate of change of the amount of substance depends on the rate it enters and leaves the tank.
- Motion Problems: Describing the velocity and position of an object under the influence of forces, where force is related to the rate of change of momentum (or mass times acceleration, $\frac{dv}{dt}$). Air resistance, for instance, is often modeled as proportional to velocity.
- Electrical Circuits: Analyzing simple circuits containing resistors (R) and inductors (L) or resistors and capacitors (C). The relationship between voltage, current, resistance, inductance, and capacitance involves derivatives with respect to time, leading to linear first-order differential equations.
- Chemical Kinetics: Modeling the rate of chemical reactions, which often depend on the concentration of reactants.
- Spread of Diseases: Simple epidemic models (like the SI or SIS models) use differential equations to describe the rates at which individuals move between categories (e.g., Susceptible to Infected).
Examples
Example 1. The population of a town grows at a rate proportional to its population. If the population was 1,00,000 in the year 2000 and 1,20,000 in 2010, estimate the population in 2020.
Answer:
Let $P(t)$ be the population of the town at time $t$ years. The growth is proportional to the population, so the differential equation is $\frac{dP}{dt} = kP$.
The general solution to this equation is $P(t) = A e^{kt}$, where $A$ and $k$ are constants.
Let $t=0$ represent the year 2000. Given: Population in 2000 ($t=0$) is $P(0) = 1,00,000$.
$P(0) = A e^{k \times 0} = A e^0 = A \times 1 = A$
$1,00,000 = A$
... (1)
So, the population model becomes $P(t) = 100000 \, e^{kt}$.
Given: Population in 2010 ($t = 2010 - 2000 = 10$ years) is $P(10) = 1,20,000$.
$P(10) = 100000 \, e^{k \times 10}$
$1,20,000 = 100000 \, e^{10k}$
Divide by $100000$:
$\frac{120000}{100000} = e^{10k}$
$1.2 = e^{10k}$
... (2)
We need to estimate the population in 2020, which corresponds to $t = 2020 - 2000 = 20$ years. We need to find $P(20)$.
$P(20) = 100000 \, e^{k \times 20}$
We can rewrite $e^{20k}$ using equation (2):
$e^{20k} = e^{2 \times 10k} = (e^{10k})^2$
[Using exponent rule $(a^m)^n = a^{mn}$]
Substitute the value from (2):
$e^{20k} = (1.2)^2 = 1.44$
... (3)
Now substitute this value back into the expression for $P(20)$:
$P(20) = 100000 \times (1.44)$
$P(20) = 1,44,000$
... (4)
Thus, the estimated population in the year 2020 is 1,44,000.
Alternatively, we could find the value of $k$ from equation (2) using logarithms: $\log(1.2) = 10k$ $k = \frac{\log(1.2)}{10} \approx \frac{0.1823}{10} = 0.01823$ Then $P(20) = 100000 e^{20 \times 0.01823} = 100000 e^{0.3646}$. Using $e^{0.3646} \approx 1.44$, we get $P(20) \approx 100000 \times 1.44 = 1,44,000$. The first method (using $e^{10k}$) is more precise as it avoids calculating $k$ directly and rounding it.
Example 2. A cup of hot tea at $80^\circ$C is placed in a room whose temperature is $25^\circ$C. If the tea cools to $50^\circ$C in 10 minutes, find the temperature of the tea after 20 minutes.
Answer:
This problem can be solved using Newton's Law of Cooling. Let $T(t)$ be the temperature of the tea at time $t$ (in minutes). The ambient temperature is $T_m = 25^\circ$C. The differential equation governing the cooling process is $\frac{dT}{dt} = k(T - T_m)$. Substituting $T_m = 25$, we get $\frac{dT}{dt} = k(T - 25)$.
The general solution to this differential equation is $T(t) = T_m + A e^{kt}$.
$T(t) = 25 + A e^{kt}$
... (1)
Given: Initial temperature at $t=0$ is $T(0) = 80^\circ$C.
$T(0) = 25 + A e^{k \times 0}$
$80 = 25 + A e^0 = 25 + A$
$A = 80 - 25 = 55$
... (2)
Substitute the value of $A$ into equation (1):
$T(t) = 25 + 55 e^{kt}$
... (3)
Given: The tea cools to $50^\circ$C in 10 minutes. So, $T(10) = 50$.
$T(10) = 25 + 55 e^{k \times 10}$
$50 = 25 + 55 e^{10k}$
Subtract 25 from both sides:
$50 - 25 = 55 e^{10k}$
$25 = 55 e^{10k}$
Divide by 55:
$\frac{25}{55} = e^{10k}$
Simplify the fraction:
$\frac{\cancel{25}^{5}}{\cancel{55}_{11}} = e^{10k}$
$\frac{5}{11} = e^{10k}$
... (4)
We need to find the temperature after 20 minutes, i.e., $T(20)$.
$T(20) = 25 + 55 e^{k \times 20} = 25 + 55 e^{20k}$
We can rewrite $e^{20k}$ using equation (4):
$e^{20k} = e^{2 \times 10k} = (e^{10k})^2$
Substitute the value from (4):
$e^{20k} = \left(\frac{5}{11}\right)^2 = \frac{25}{121}$
... (5)
Now substitute this value back into the expression for $T(20)$:
$T(20) = 25 + 55 \left(\frac{25}{121}\right)$
Simplify the expression:
$T(20) = 25 + \cancel{55}^{5} \times \frac{25}{\cancel{121}_{11}}$
$T(20) = 25 + \frac{5 \times 25}{11} = 25 + \frac{125}{11}$
Combine the terms:
$T(20) = \frac{25 \times 11}{11} + \frac{125}{11} = \frac{275}{11} + \frac{125}{11} = \frac{275 + 125}{11} = \frac{400}{11}$
$T(20) = \frac{400}{11}^\circ\text{C}$
... (6)
As a decimal, $\frac{400}{11} \approx 36.36^\circ$C.
The temperature of the tea after 20 minutes is $\frac{400}{11}^\circ$C or approximately $36.36^\circ$C.