Linear Differential Equations: Definition, Form, Methods & Examples  

1.0Introduction to Linear Differential Equations

Linear differential equations are a crucial topic for the JEE Main and Advanced examinations. Mastering this concept is essential as questions from this section are frequently asked and are generally scoring. A first-order linear differential equation is of the form:

$\frac{dy}{dx}+P(x)y=Q(x)$

where P(x) and Q(x) are constants or continuous functions of x. The first step in recognizing and resolving these equations is comprehending this form. Once the given equation has been rearranged into this standard format, the key is to correctly identify P(x) and Q(x).

2.0What is a Linear Differential Equation?

A differential equation is one that has an unknown function and its derivatives. A linear differential equation is one in which the unknown function and its derivatives only show up in the first degree, and there are no products of the unknown function and its derivatives. We mostly study first-order linear differential equations for the JEE. The "linearity" comes from the fact that the variables and their derivatives are raised to the first power.

3.0General Form & Identification

The standard form of a first-order linear differential equation is:

$\frac{dy}{dx}+P(x)y=Q(x)$

Here,

• $\frac{dy}{dx}$ is the first derivative of y with respect to x.
• P(x) is a function of x (or a constant) and is the coefficient of y.
• Q(x) is a function of x (or a constant) and is the term on the right-hand side of the equation.

Important: Before you can apply the solution method, always rearrange the given equation to match this standard form.

4.0Step-by-Step Method to Solve Linear Differential Equations

Solving a linear differential equation is a systematic three-step process.

Step 1: The Standard Form

Ensure the equation is in the form $\frac{dy}{dx}+P(x)y=Q(x)$ . If the coefficient of $\frac{dy}{dx}$ is not 1, divide the entire equation by that coefficient.

Step 2: Finding the Integrating Factor (I.F.)

The integrating factor (I.F.) is a crucial term that simplifies the equation. It is calculated using the formula:

$\text{I.F.}=e^{\int P(x)dx}$

The integration here is simple, and you don't need to add the constant of integration.

Step 3: The General Solution

Once you have the I.F., the general solution to the differential equation is given by the formula:

y⋅(I.F.)=∫Q(x)⋅(I.F.)dx+C

where C is the constant of integration. This formula directly gives the solution for y in terms of x.

5.0Working with the Integrating Factor

The integrating factor is a function that, when multiplied throughout the linear differential equation, makes the left-hand side a perfect derivative of the product of the dependent variable (y) and the integrating factor itself.

Formula for I.F.

$\text{I.F.}=e^{\int P(x)dx}$

Example Calculation

Consider the equation: $\frac{dy}{dx}+y\cot (x)=\sin (x)$

Here, P(x)=cot(x).

The integral of P(x) is ∫cot(x)dx=ln∣sin(x)∣.

So, the integrating factor is:

$\text{I.F.}=e^{\int \cot (x)dx}=e^{\ln |\sin (x)|}=|\sin (x)|$

For the JEE, we can usually assume sin(x) is positive in the given domain, so we use sin(x).

6.0Solved Examples on Linear Differential Equations (JEE-Level)

Example 1: Basic Application

Problem: Solve the differential equation $x\frac{dy}{dx}+y=x^3$

Solution:

1. Standard Form: Divide by x:
   $\frac{dy}{dx}+\frac{1}{x}y=x^2$
   
   Comparing with dy/dx ​+P(x)y=Q(x), we get $P(x)=x_{1}andQ(x)=x_2$.
2. Integrating Factor: $I.F.=e^{\int \frac{1}{x}dx}=e^{\ln (x)}=x$
3. General Solution:

$$\begin{gathered} y\cdot (x)=\int x^2\cdot (x)dx+C \\ xy=\int x^{3}dx+C \\ xy=\frac{x^4}{4}+C \end{gathered}$$
The final solution is $y=\frac{x^2}{4}+\frac{C}{x}$

Example 2: A Slightly More Complex Problem

Problem: Solve $\frac{dy}{dx}+\frac{2x}{1+x^2}y=\frac{1}{1+x^2}$

Solution:

1. Standard Form: The equation is already in the standard form. Here, ​ and $P(x)=\frac{2x}{1+x^2}\text{and}Q(x)=\frac{1}{1+x^2}$.

Integrating Factor:
I.F. = $e^{\int \frac{2x}{1+x^2}dx}$
Let so dt=2xdx.

$$\begin{gathered} \int \frac{2x}{1+x^2}dx=\int \frac{dt}{t}=\ln |t|=\ln (1+x^2) \\ I.F.=e^{\ln (1+x^2)}=1+x^2 \end{gathered}$$

2. General Solution:

$$\begin{gathered} y\cdot (1+x^2)=\int \frac{1}{1+x^2}\cdot (1+x^2)dx+C \\ y(1+x^2)=\int 1dx+C \\ y(1+x^2)=x+C \end{gathered}$$
Final solution: $y=\frac{x+C}{1+x^2}$

Example 3: Non-Standard Form

Problem: Solve $\sin (x)\frac{dy}{dx}+y\cos (x)=x\sin (x)$.

Solution:

1. Standard Form: Divide the entire equation by sin(x) to make the coefficient of $\frac{dy}{dx}$ equal to 1.

$\frac{dy}{dx}+y\frac{\cos (x)}{\sin (x)}=x$

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

Now, P(x)=cot(x) and Q(x)=x.

2. Integrating Factor:
   I.F. = $e^{\int \cot (x)dx}=e^{\ln |\sin (x)|}=\sin (x)$
   General Solution:

$y\cdot \sin (x)=\int x\cdot \sin (x)dx+C$
Use integration by parts for ∫xsin(x)dx:

$\int x\sin (x)dx=x(-\cos (x))-\int 1\cdot (-\cos (x))dx$

$=-x\cos (x)+\int \cos (x)dx$

$=-x\cos (x)+\sin (x)$
So, ysin(x)=−xcos(x)+sin(x)+C.
Final solution: $y=\frac{-x\cos (x)+\sin (x)+C}{\sin (x)}=-x\cot (x)+1+C\csc (x)$

7.0Alternative Form: Equation in x and y

A linear differential equation can also be written in the form:

$\frac{dx}{dy}+P(y)x=Q(y)$

Here, P(y) and Q(y) are functions of y (or constants).

The solution method remains the same, but with a slight change:

• Integrating Factor: $\text{I.F.}=e^{\int P(y)dy}$
• General Solution: x⋅(I.F.)=∫Q(y)⋅(I.F.)dy+C 

8.0Key Tips & Tricks for JEE

• Recognition is Key: The first and most important step is to correctly identify the equation as a linear differential equation. Look for the standard form $\frac{dy}{dx}+P(x)y=Q(x)$.
• Coefficient of $\frac{dy}{dx}$ : Always ensure the coefficient of $\frac{dy}{dx}$ is 1 before finding P(x) and Q(x).
• Integration Skills: A strong grasp of integration, especially methods like integration by parts and substitution, is crucial for solving these problems.
• Practice with Multiple Forms: Practice problems where the equation is not in the standard form and requires manipulation. This is a common trick in JEE questions.

9.0Common Mistakes to Avoid

• Not converting to the standard form: This is the most frequent error. Always divide by the coefficient of $\frac{dy}{dx}$ first.
• Incorrectly identifying P(x) and Q(x): Make sure you correctly identify the function that is the coefficient of y and the function on the right-hand side.
• Errors in Integration: Be careful while calculating the integrating factor and the final integral. A small mistake can lead to a completely wrong answer.
• Forgetting the Constant of Integration (C): The constant of integration is essential for a general solution. Don't forget to add it at the end. For problems with initial conditions, use this constant to find the particular solution.

10.0Practice Problems on Linear Differential Equations

1. Solve: $(1+x^2)\frac{dy}{dx}+2xy=\cos (x)$

2. Solve: $x\ln (x)\frac{dy}{dx}+y=2\ln (x)$

3. Solve: $\frac{dy}{dx}+y\tan (x)=\sec (x)$

4. Solve: $\frac{dx}{dy}+\frac{x}{y}=y^2(\text{This is an example of the alternative form!})$