Let v be the solution of the differential equation $\left(1-x^{2}\right) d y=\left(x y+\left(x^{3}+2\right) \sqrt{1-x^{2}}\right) d x,-1<x<1$ and $y(0)=0$ if $\int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{1-x^{2}} y(x) d x=k$ then $k^{-1}$ is equal to :

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Differential Equations

Let v be the solution of the differential equation

(1 - x^{2}) d y = (x y + (x^{3} + 2) 1 - x^{2}) d x, - 1 < x < 1

and

y (0) = 0

\int_{- \frac{1}{2}}^{\frac{1}{2}} 1 - x^{2} y (x) d x = k

then

k^{- 1}

is equal to :

Answer: 320

Rewriting the differential equation:

\frac{d y}{d x} + \frac{- x}{1 - x ^{2}} y = \frac{x ^{3} + 2}{1 - x ^{2}}

This is a first-order linear ODE. The integrating factor is:

I F = e^{\int \frac{- x}{1 - x ^{2}} d x} = 1 - x^{2}

Multiplying through and integrating:

y 1 - x^{2} = \int \frac{x ^{3} + 2}{1 - x ^{2}} \cdot 1 - x^{2} d x = \int (x^{3} + 2) d x = \frac{x ^{4}}{4} + 2 x + C

Using

y (0) = 0

C = 0

, so

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

.

The integral becomes:

k = \int_{- 1/2}^{1/2} (\frac{x ^{4}}{4} + 2 x) d x

Since

2 x

is odd, its integral over a symmetric interval is 0:

k = \frac{1}{4} \cdot 2 \int_{0}^{1/2} x^{4} d x = \frac{1}{2} \cdot \frac{1}{5} [\frac{1}{32}] = \frac{1}{320}

Therefore

k^{- 1} = 320

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Integer answer