PYQ Vault

JEE Mains Maths · Differential Equations

Linear Equations: The Integrating Factor

Solving dy/dx + P(x)y = Q(x) by multiplying through by the integrating factor, the exponential of the integral of P, which turns the left side into the derivative of a product.

Why this matters

Twenty-seven PYQs, ten of them numerical answer. Each is one routine: write the equation in standard form, find the integrating factor, integrate, and fix the constant. The two ideas differ only in where the factor comes from — algebraic P, or trigonometric and inverse-trigonometric P.

Concept 1 of 2: Standard form and the integrating factor

Divide by the coefficient of dydx\frac{dy}{dx} to reach dydx+Py=Q\frac{dy}{dx}+Py=Q. The integrating factor is the multiplier μ=e∫P dx\mu=e^{\int P\,dx}; it is chosen so that μy′+μPy=(μy)′\mu y'+\mu Py=(\mu y)'. Then μy=∫μQ dx+C\mu y=\int\mu Q\,dx+C. With P=kxP=\frac kx, μ=xk\mu=x^k; with P=2x1+x2P=\frac{2x}{1+x^2}, μ=1+x2\mu=1+x^2.

Definition

  • Standard form: dydx+P(x) y=Q(x)\frac{dy}{dx}+P(x)\,y=Q(x).
  • Integrating factor μ=e∫P dx\mu=e^{\int P\,dx}; then μy=∫μQ dx+C\mu y=\int\mu Q\,dx+C.
  • e∫Pe^{\int P} for P=kxP=\frac kx: xkx^k; for P=1xln⁡xP=\frac{1}{x\ln x}: ln⁡x\ln x; for P=−x1−x2P=-\frac{x}{1-x^2}: 1−x2\sqrt{1-x^2}.
  • Constant P=kP=k: μ=ekx\mu=e^{kx}.

Linear equation

dydx+Py=Q ⇒ y e∫P dx=∫Q e∫P dx dx+C\frac{dy}{dx}+Py=Q\ \Rightarrow\ y\,e^{\int P\,dx}=\int Q\,e^{\int P\,dx}\,dx+C

Worked example

Solve xdydx+2y=x2x\frac{dy}{dx}+2y=x^2 with y(1)=1y(1)=1, and find y(2)y(2).
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 26 June 2022 · Q85Moderate

Example 1 · Differential Equations · Linear Equations: The Integrating Factor

Let the solution curve y=y(x)y=y(x) of the differential equation (4+x2)dy−2x(x2+3y+4)dx=0\left( 4 +x^{2} \right)dy- 2x\left( x^{2}+ 3y+ 4 \right)dx= 0 pass through the origin. Then y(2)y(2) is equal to

Standard form before the factor

The integrating factor comes from PP only after the coefficient of y′y' is 1. In x y′+2y=x2x\,y'+2y=x^2, PP is 2x\frac2x, not 2.

Concept 2 of 2: Integrating factors from trigonometric P

The same routine, with the integrals that make trigonometric factors. P=tan⁡xP=\tan x gives sec⁡x\sec x; P=2tan⁡xP=2\tan x gives sec⁡2x\sec^2x; P=−tan⁡xP=-\tan x gives cos⁡x\cos x. P=11+x2P=\frac{1}{1+x^2} gives etan⁡−1xe^{\tan^{-1}x}, and then putting t=tan⁡−1xt=\tan^{-1}x makes ∫μQ dx\int\mu Q\,dx routine.

Definition

  • e∫tan⁡x dx=sec⁡xe^{\int\tan x\,dx}=\sec x; e∫2tan⁡x dx=sec⁡2xe^{\int2\tan x\,dx}=\sec^2x; e−∫tan⁡x dx=cos⁡xe^{-\int\tan x\,dx}=\cos x.
  • e∫sec⁡2x dx=etan⁡xe^{\int\sec^2x\,dx}=e^{\tan x}; e∫cos⁡x dx=esin⁡xe^{\int\cos x\,dx}=e^{\sin x}.
  • e∫dx1+x2=etan⁡−1xe^{\int\frac{dx}{1+x^2}}=e^{\tan^{-1}x}; then put t=tan⁡−1xt=\tan^{-1}x.
  • ∫sec⁡xtan⁡x dx=sec⁡x\int\sec x\tan x\,dx=\sec x.

Common trigonometric factors

P=ktan⁡x ⇒ μ=sec⁡kx,P=11+x2 ⇒ μ=etan⁡−1xP=k\tan x\ \Rightarrow\ \mu=\sec^kx,\qquad P=\frac{1}{1+x^2}\ \Rightarrow\ \mu=e^{\tan^{-1}x}

Worked example

Solve dydx+ytan⁡x=cos⁡x\frac{dy}{dx}+y\tan x=\cos x with y(0)=0y(0)=0, and find y(π3)y\left(\frac\pi3\right).
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 6 April 2024 · Q77Moderate

Example 2 · Differential Equations · Linear Equations: The Integrating Factor

Let y=y(x)y = y(x) be the solution of the differential equation (1+x2)dydx+y=etan⁡−1x,y(1)=0\left( 1 +x^{2} \right)\frac{dy}{dx}+ y =e^{\tan^{- 1}x},y(1) = 0. Then y(0)y(0) is

Sign of the tangent term

∫tan⁡x dx=ln⁡sec⁡x\int\tan x\,dx=\ln\sec x, so +tan⁡x+\tan x gives sec⁡x\sec x and −tan⁡x-\tan x gives cos⁡x\cos x. Swapping them is the commonest slip on this page.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Standard form and the integrating factor

    Linear equation

    dydx+Py=Q ⇒ y e∫P dx=∫Q e∫P dx dx+C\frac{dy}{dx}+Py=Q\ \Rightarrow\ y\,e^{\int P\,dx}=\int Q\,e^{\int P\,dx}\,dx+C
  • Integrating factors from trigonometric P

    Common trigonometric factors

    P=ktan⁡x ⇒ μ=sec⁡kx,P=11+x2 ⇒ μ=etan⁡−1xP=k\tan x\ \Rightarrow\ \mu=\sec^kx,\qquad P=\frac{1}{1+x^2}\ \Rightarrow\ \mu=e^{\tan^{-1}x}

Watch out for (2)

Test yourself on Differential Equations

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.