Question

If $y=\frac{x{\sin }^{-1}x}{\sqrt{1-x^2}},$ prove that $(1-x^2)\frac{dy}{dx}=x+\frac{y}{x}$.

Answer 1

We have,

$y=\frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}$

On differentiating w.r.t $x$, we get

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

Hence, proved.