The function y = f(x) is the solution of the differential equation dy/dx + xy/(x^2-1) = fracx^4+2x√(1-x^2) in (-1,1) satisfying f(0) = 0. Then int_-frac√(3)2^frac√(3)2 f(x) dx is

The function y = f(x) is the solution of the differential… — JEE Main 2014 Mathematics PYQ

The function

y = f(x)

is the solution of the differential equation

\frac{dy}{dx} + \frac{xy}{x^2-1} = \frac{x^4+2x}{\sqrt{1-x^2}}

(-1,1)

satisfying

f(0) = 0

. Then

\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x)\, dx

A.
$\frac{\pi}{3} - \frac{\sqrt{3}}{2}$
B.
$\frac{\pi}{3} - \frac{\sqrt{3}}{4}$
C.
$\frac{\pi}{6} - \frac{\sqrt{3}}{4}$
D.
$\frac{\pi}{6} - \frac{\sqrt{3}}{2}$

Show answer and solution

Correct option: B

Solution

The integrating factor is e^(∫x/(x²-1)dx) = e^((1/2)ln|x²-1|) = √(1-x²) (taking the appropriate sign in (-1,1)).

Multiplying through: d/dx[y√(1-x²)] = x⁴+2x)/√(1-x²) · √(1-x²)... wait, let me redo. The equation becomes d/dx[f(x)·√(1-x²)] = x⁴+2x (after multiplying both sides by √(1-x²)), so integrating: f(x)·√(1-x²) = x⁵/5 + x² + C. With f(0)=0, C=0, giving f(x) = (x⁵/5 + x²)/√(1-x²).

Now split the integral from -√3/2 to √3/2: the x⁵/5·(1/√(1-x²)) part is odd, so it vanishes. The remaining part is 2∫₀^(√3/2) x²/√(1-x²) dx.

Substitute x = sinθ: 2∫₀^(π/3) sin²θ dθ = 2·[θ/2 - sin2θ/4]₀^(π/3) = [θ - sin2θ/2]₀^(π/3) = π/3 - sin(2π/3)/2 = π/3 - (√3/2)/2 = **π/3 - √3/4

[JEE Main 2014]

JEE Main 2014 Mathematics question on Differential Equations