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Differential Equations appeared 46 times across 3 years — 5.3% of Mathematics. This question is from Linear Differential Equations.

Year 2026 2025 2024 Total
Questions 13 17 16 46

Let y = y(x) be the solution of the differential equation (x² + 1)y' - 2xy = (x⁴ + 2x² + 1) x, y(0) = 1. Then ∫₋₃³ y(x) dx is:

Solution & Explanation

Related Formula

For a linear differential equation (dy)/(dx) + Py = Q, the Integrating Factor (IF) is defined as:

IF = e∫ P dx
Core Logic

Divide the full differential equation by (x²+1):

(dy)/(dx) - ((2x)/(x²+1))y = ((x²+1)² x)/(x²+1) = (x²+1) x

This is a standard Linear Differential Equation with:

P = -(2x)/(x²+1), Q = (x²+1) x IF = e∫ -(2x)/(x²+1) dx = e-ln(x²+1) = (1)/(x²+1)
Step 1: Solve for General Solution

The solution format is y · IF = ∫ Q · IF dx:

y · (1)/(x²+1) = ∫ (x²+1) x · (1)/(x²+1) dx (y)/(x²+1) = x + c

Using the boundary condition y(0) = 1:

(1)/(0+1) = (0) + c c = 1 y = (x²+1)(sin x + 1)
Step 2: Definite Integration Evaluation

We need to evaluate ∫₋₃³ y dx:

∫₋₃³ (x²+1)(sin x + 1) dx = ∫₋₃³ (x² x + x² + x + 1) dx

By symmetry of odd/even functions over symmetric intervals [-a, a]: ∫₋₃³ x² x dx = 0 (since it is an odd function) ∫₋₃³ x dx = 0 (since it is an odd function)

Thus, we are left with the even components:

∫₋₃³ (x² + 1) dx = 2 ∫₀³ (x² + 1) dx = 2 [ (x³)/(3) + x ]₀³ = 2(9 + 3) = 24
Pattern Recognition

Splitting a symmetric interval integral into odd and even parts immediately simplifies calculations by dropping all odd functions down to zero.

Chapter Mix

Class 12 Mathematics: Differential Equations Class 12 Mathematics: Integral Calculus

Reference Study Guides

More Differential Equations Previous-Year Questions — Page 10

Q11 jee_main_2024_31_jan_morning Linear Differential Equations
Let y = y(x) be the solution of the differential equation (dy)/(dx) = (( x) + y)/( x( x - x x)), x in (0, (π)/(2)) satisfying the condition y((π)/(4)) = 2. Then, y((π)/(3)) is
  • A. √(3)(2 + ₑ√(3))
  • B. √(3)2(2 + ₑ 3)
  • C. √(3)(1 + 2 ₑ 3)
  • D. √(3)(2 + ₑ 3)

Solution

Core Logic
(dy)/(dx) = (( x)/( x) + y)/( x ((1)/( x) - ( ² x)/( x))) = ( x + y x)/( x (1 - ² x)) (dy)/(dx) = ( x + y x)/( x ² x) = ² x + (2y)/( 2x) (dy)/(dx) - 2 (2x)y = ² x
Step 1: Integrating Factor

This is an LDE of form (dy)/(dx) + Py = Q.

I.F. = e∫ -2 (2x) dx

Let 2x = t 2dx = dt.

I.F. = e-∫ t dt = e-ln| (t/2)| = e-ln| x| = (1)/(| x|)
Step 2: Solution of LDE
y(I.F.) = ∫ Q(I.F.) dx + C y(1)/( x) = ∫ ² x (1)/( x) dx + C

Let x = t ² x dx = dt.

y(1)/( x) = ∫ (dt)/(t) + C = ln| x| + C y = x(ln| x| + C)
Step 3: Boundary Value

Given y(π/4) = 2:

2 = 1(ln 1 + C) C = 2

Thus, y = x (ln| x| + 2). At x = π/3:

y(π/3) = √(3)(ln√(3) + 2)
Chapter Mix

Class 12 Maths: Differential Equations

More Differential Equations Questions — jee_main_2025_07_april_evening

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