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Linear Differential Equations and Integrating Factor

A first-order linear differential equation in y has the form dy/dx + P(x)y = Q(x), where P and Q are functions of x. It is solved using the integrating factor e^(∫P(x) dx).

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Student-friendly explanation

The integrating factor changes the left side into the derivative of y multiplied by the integrating factor. After multiplying by IF, the equation becomes d/dx(y·IF) = Q·IF. Integrating both sides gives y·IF = ∫Q·IF dx + C.

How to write this in exams

  1. 1

    Start with the exact idea

    A first-order linear differential equation in y has the form dy/dx + P(x)y = Q(x), where P and Q are functions of x. It is solved using the integrating factor e^(∫P(x) dx).

  2. 2

    Then show how to use it

    Rewrite the equation so the coefficient of dy/dx is 1. Identify P(x) and Q(x). Find IF = e^(∫P dx). Multiply the whole equation by IF. Write the left side as d(y·IF)/dx. Integrate the right side Q·IF. Add C and solve for y if required.

  3. 3

    Add one concrete example

    For dy/dx + y = e^x, P = 1 and Q = e^x. IF = e^(∫1 dx) = e^x. Then y e^x = ∫e^x·e^x dx + C = ∫e^(2x) dx + C = e^(2x)/2 + C.

  4. 4

    Avoid this incomplete answer

    Taking P before dividing by the coefficient of dy/dx, or forgetting to multiply Q by the integrating factor before integration.

Definition

A first-order linear differential equation in y has the form dy/dx + P(x)y = Q(x), where P and Q are functions of x. It is solved using the integrating factor e^(∫P(x) dx).

Example

For dy/dx + y = e^x, P = 1 and Q = e^x. IF = e^(∫1 dx) = e^x. Then y e^x = ∫e^x·e^x dx + C = ∫e^(2x) dx + C = e^(2x)/2 + C.

Rule to remember

Standard form: dy/dx + P(x)y = Q(x). Condition: y and dy/dx occur to the first power, and P, Q are functions of x only. Integrating factor: IF = e^(∫P dx). Solution rule: y·IF = ∫Q·IF dx + C. For dx/dy + P(y)x = Q(y), use IF = e^(∫P(y) dy).

Memory hook

Linear form first, IF from P, then y times IF equals integral of Q times IF.

Examples and method

Worked example

Example: Solve dy/dx + (1/x)y = x, x > 0. Here P = 1/x and Q = x. IF = e^(∫1/x dx) = e^(log x) = x. Multiply by IF: x(dy/dx) + y = x2. Since left side is d(xy)/dx, d(xy)/dx = x2. Integrate: xy = x3/3 + C. Hence y = x2/3 + C/x.

Method to apply

Rewrite the equation so the coefficient of dy/dx is 1. Identify P(x) and Q(x). Find IF = e^(∫P dx). Multiply the whole equation by IF. Write the left side as d(y·IF)/dx. Integrate the right side Q·IF. Add C and solve for y if required.

Diagram support

A diagram is not required for standard linear differential equation solving. If a case-study question describes growth, decay, or motion, a table or graph may support interpretation but is not essential to the method.

How CBSE asks it

Often appears as a long-answer problem or as part of an application-based question. The equation may need rearrangement before it matches dy/dx + Py = Q.

Avoid common mistakes

Common confusion

Students sometimes use e^(∫Q dx) as the integrating factor. The integrating factor depends on P, not Q, after the equation is in dy/dx + Py = Q form.

Common wrong answer

Taking P before dividing by the coefficient of dy/dx, or forgetting to multiply Q by the integrating factor before integration.

Exam tip

First divide by the coefficient of dy/dx so that its coefficient becomes 1. Only then identify P and Q and calculate the integrating factor.

Quick check

What is the integrating factor of dy/dx + (2/x)y = x3, for x > 0?

Here P = 2/x, so IF = e^(∫2/x dx) = e^(2 log x) = x2.

Answer writing and exam use

1-mark answer

A first-order linear differential equation in y has the form dy/dx + P(x)y = Q(x), where P and Q are functions of x. It is solved using the integrating factor e^(∫P(x) dx).

2-mark answer

A first-order linear differential equation in y has the form dy/dx + P(x)y = Q(x), where P and Q are functions of x. It is solved using the integrating factor e^(∫P(x) dx). Standard form: dy/dx + P(x)y = Q(x). Condition: y and dy/dx occur to the first power, and P, Q are functions of x only. Integrating factor: IF = e^(∫P dx). Solution rule: y·IF = ∫Q·IF dx + C. For dx/dy + P(y)x = Q(y), use IF = e^(∫P(y) dy). For dy/dx + y = e^x, P = 1 and Q = e^x. IF = e^(∫1 dx) = e^x. Then y e^x = ∫e^x·e^x dx + C = ∫e^(2x) dx + C = e^(2x)/2 + C.

3-mark answer

The integrating factor changes the left side into the derivative of y multiplied by the integrating factor. After multiplying by IF, the equation becomes d/dx(y·IF) = Q·IF. Integrating both sides gives y·IF = ∫Q·IF dx + C. Standard form: dy/dx + P(x)y = Q(x). Condition: y and dy/dx occur to the first power, and P, Q are functions of x only. Integrating factor: IF = e^(∫P dx). Solution rule: y·IF = ∫Q·IF dx + C. For dx/dy + P(y)x = Q(y), use IF = e^(∫P(y) dy). Example: Solve dy/dx + (1/x)y = x, x > 0. Here P = 1/x and Q = x. IF = e^(∫1/x dx) = e^(log x) = x. Multiply by IF: x(dy/dx) + y = x2. Since left side is d(xy)/dx, d(xy)/dx = x2. Integrate: xy = x3/3 + C. Hence y = x2/3 + C/x. Often appears as a long-answer problem or as part of an application-based question. The equation may need rearrangement before it matches dy/dx + Py = Q. Taking P before dividing by the coefficient of dy/dx, or forgetting to multiply Q by the integrating factor before integration.
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