The function $y=f(x)$ is the solution of the differential equation \[ \frac{d y}{d x}+\frac{x y}{x^{2}-1}=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}} \] in $(-1,1)$ satisfying $f(0)=0$. Then \[ \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) d x \] is

(B) $\frac{\pi}{3}-\frac{\sqrt{3}}{4}$

(A) $\frac{\pi}{3}-\frac{\sqrt{3}}{2}$

(C) $\frac{\pi}{6}-\frac{\sqrt{3}}{4}$

(D) $\frac{\pi}{6}-\frac{\sqrt{3}}{2}$

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Differential Equations

The function

y = f (x)

is the solution of the differential equation

\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{4} + 2 x}{1 - x ^{2}}

(- 1, 1)

satisfying

f (0) = 0

. Then

\int_{- \frac{3}{2}}^{\frac{3}{2}} f (x) d x

Answer: (B)

The given differential equation is linear. The integrating factor is

e^{\int \frac{x}{x ^{2} - 1} d x} = e^{\frac{1}{2} l o g ∣1 - x^{2} ∣} = 1 - x^{2}

.

Multiplying both sides by the integrating factor:

y 1 - x^{2} = \int \frac{x ^{4} + 2 x}{1 - x ^{2}} \cdot 1 - x^{2} d x = \int (x^{4} + 2 x) d x = \frac{x ^{5}}{5} + x^{2} + c

Using

f (0) = 0

, we get

c = 0

. Thus

y = \frac{x ^{5} + 5 x ^{2}}{5 1 - x ^{2}}

.

Now,

\int_{- 3 /2}^{3 /2} f (x) d x = \int_{- 3 /2}^{3 /2} \frac{x ^{5}}{5 1 - x ^{2}} d x + \int_{- 3 /2}^{3 /2} \frac{x ^{2}}{1 - x ^{2}} d x

.

The first integral is zero (odd function), and the second integral evaluates to

\frac{π}{3} - \frac{3}{4}

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Single correct option