Let $y=f(x)$ be the solution of the differential equation $\frac{d y}{d x}+\frac{x y}{x^{2}-1}=\frac{x^{6}+4 x}{\sqrt{1-x^{2}}},-1<x<1$ such that $f(0)=0$. If $6 \int_{-1 / 2}^{1 / 2} f(x) d x=2 \pi-\alpha$ then $\alpha^{2}$ is equal to ____.

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Differential Equations

Let

y = f (x)

be the solution of the differential equation

\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{6} + 4 x}{1 - x ^{2}}, - 1 < x < 1

such that

f (0) = 0

. If

6 \int_{- 1/2}^{1/2} f (x) d x = 2 π - α

then

α^{2}

is equal to ____.

Answer: 27

The equation

\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{6} + 4 x}{1 - x ^{2}}

is linear, with integrating factor

I = exp (\int \frac{x}{x ^{2} - 1} d x) = 1 - x^{2} (since x^{2} - 1 < 0 on (- 1, 1)) .

Multiplying through:

\frac{d}{d x} (y 1 - x^{2}) = x^{6} + 4 x,

y 1 - x^{2} = \frac{x ^{7}}{7} + 2 x^{2} + C .

f (0) = 0

gives

C = 0

, so

f (x) = \frac{x ^{7} /7 + 2 x ^{2}}{1 - x ^{2}}

.

In

6 \int_{- 1/2}^{1/2} f (x) d x

, the

x^{7}

-term is odd (denominator is even) and integrates to zero over the symmetric interval, leaving

6 \int_{- 1/2}^{1/2} \frac{2 x ^{2}}{1 - x ^{2}} d x = 24 \int_{0}^{1/2} \frac{x ^{2}}{1 - x ^{2}} d x .

Put

x = sin θ

: the limits become

0

π /6

, and

24 \int_{0}^{π /6} sin^{2} θ d θ = 12 [θ - \frac{sin 2 θ}{2}]_{0}^{π /6} = 12 (\frac{π}{6} - \frac{3}{4}) = 2 π - 33 .

Comparing with

2 π - α

α = 33

, so

α^{2} = 27.

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Integer answer