Question

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 $\int\limits_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{1-x^{2}} y(x) d x=k$ then $k^{-1}$ is equal to :

Answer 1

$\left(1-x^{2}\right) \frac{d y}{d x}=x y+\left(x^{3}+2\right) \sqrt{1-x^{2}}$
$\Rightarrow \frac{d y}{d x}+\left(\frac{-x}{1-x^{2}}\right) y=\frac{x^{3}+2}{\sqrt{1-x^{2}}}$
$I F=e^{\int \frac{-x}{1-x^{2}} d x}=\sqrt{1-x^{2}}$
$y(x) \cdot \sqrt{1-x^{2}}=\frac{x^{4}}{4}+2 x+c$
$y(0)=0 \Rightarrow c=0$
$\sqrt{1-x^{2}} y (x)=\frac{x^{4}}{4}+2 x$
required value $=\int\limits_{-1 / 2}^{1 / 2}\left(\frac{x^{4}}{4}+2 x\right) d x-\frac{1}{4} \cdot 2 \int\limits_{0}^{1 / 2} x^{4} d x$
$=\frac{1}{10}\left(x^{5}\right)_{0}^{1 / 2}=\frac{1}{320}$
$k ^{-1}=320$