Question

Let $v$ be the solution of the differential equation $\left(1−x^2\right)dy=\left(xy+\left(x^3+2\right)\sqrt{1−x^2}\right)dx,−1<x<1$ and $y(0)=0$ if $\underset{−\frac{1}{2}}{\overset{\frac{1}{2}}{∫}}\sqrt{1−x^2}y(x)dx=k$ then $k^{−1}$ is equal to :

Answer 1

Answer: 320

$\left(1−x^2\right)\frac{dy}{dx}=xy+\left(x^3+2\right)\sqrt{1−x^2}$
$⇒\frac{dy}{dx}+\left(\frac{−x}{1−x^2}\right)y=\frac{x^3+2}{\sqrt{1−x^2}}$
$IF=e^{∫\frac{−x}{1−x^2}dx}=\sqrt{1−x^2}$
$y(x)â‹\dots \sqrt{1−x^2}=\frac{x^4}{4}+2x+c$
$y(0)=0⇒c=0$
$\sqrt{1−x^2}y(x)=\frac{x^4}{4}+2x$
required value $=\underset{−1/2}{\overset{1/2}{∫}}\left(\frac{x^4}{4}+2x\right)dx−\frac{1}{4}â‹\dots 2\underset{0}{\overset{1/2}{∫}}{x}^{4}dx$
$=\frac{1}{10}{\left(x^5\right)}_0^{1/2}=\frac{1}{320}$
$k^{−1}=320$