Let $y=y\left(x\right)$ 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 $y\left(0\right)=0$. If ${\int }_{-\frac{1}{2}}^{\frac{1}{2}}\sqrt{1-x^2}y\left(x\right)dx=k$ then $k^{-1}$ is equal to

Given,

$\left(1-x^2\right)\frac{dy}{dx}=xy+\left(x^3+2\right)\sqrt{1-x^2}$

$\Rightarrow \frac{dy}{dx}+\left(\frac{-x}{1-x^2}\right)y=\frac{x^3+2}{\sqrt{1-x^2}}$

$\mathrm{IF}={\mathrm{e}}^{\int \frac{-x}{1-x^2}dx}=\sqrt{1-x^2}$

$y\left(x\right)·\sqrt{1-x^2}=\frac{x^4}{4}+2x+c$

$y\left(0\right)=0\Rightarrow c=0$

$\sqrt{1-x^2}y\left(x\right)=\frac{x^4}{4}+2x$

So,required value $={\int }_{-\frac{1}{2}}^{\frac{1}{2}}\left(\frac{x^4}{4}+2x\right)dx-\frac{1}{4}·2{\int }_0^{\frac{1}{2}}{x}^{4}dx$

$=\frac{1}{10}{\left(x^5\right)}_0^{\frac{1}{2}}=\frac{1}{320}$

$k^{-1}=320$