JEE Main 2025 · 7 April · evening · Single correct · +4 / −1
Let y = y(x) be the solution of the differential equation (x² + 1)y' - 2xy = (x⁴ + 2x² + 1)cos x,
y(0) = 1. Then ∫₋₃³ y(x) dx is :
- (a) 36
- (b) 24
- (c) 18
- (d) 30
Answer and solution
Answer (b) 24
Solution
Maths sir · 5 steps
Divide through by x²+1 first, then the odd part of y disappears over [-3, 3].
- Standard form. Note that x⁴ + 2x² + 1 = (x²+1)². Dividing by x²+1 gives y' - 2x/(x²+1) y = (x²+1)cos x.
- Integrating factor. IF = e^(-∫ 2x/(x²+1)dx) = e^(-ln(x²+1)) = 1/(x²+1). Multiplying by it turns the left side into an exact derivative.
d/dx(y/(x²+1)) = cos x ⇒ y/(x²+1) = sin x + C
- Use y(0) = 1. At x = 0: 1/1 = 0 + C, so C = 1. Hence y = (x²+1)sin x + (x²+1).
- Use symmetry. (x²+1)sin x is odd, so its integral over [-3, 3] is 0. The even part gives 2∫₀³ (x²+1) dx = 2(9 + 3) = 24.
- So. ∫₋₃³ y(x) dx = 24, which is option (b).
The idea to keep. Before integrating over a symmetric interval like [-a, a], split the function into odd and even parts. The odd part gives 0, so you only integrate the even part.
Answer from the NTA final answer key. Solution by PrepWiser.