Question

If $y={\left({\sin }^{-1}x\right)}^2$, then prove that $(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}-2=0$

Answer 1

Given $y={\left({\sin }^{-1}x\right)}^2$

On differentiating $y$ with respect to $x$.

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

On again differentiating $\frac{dy}{dx}$ with respect to $x$.

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

$\frac{2[1+\frac{x.{\sin }^{-1}x}{\sqrt{1-x^2}}]}{\sqrt{1-x^2}}=\frac{2\sqrt{1-x^2}+2x{\sin }^{-1}x}{\sqrt{1-x^2(1-x^2)}}$

On putting value of $\frac{dy}{d}$ and $\frac{{\partial }^{2}y}{\partial x^2}$ in LHS

$(1-x^2)\frac{{\partial }^{2}y}{\partial x^2}-x\frac{dy}{dx}-2=0$

$=(1-x^2)\frac{2\sqrt{1-x^2}+2x{\sin }^{-1}x}{\sqrt{1-x^2(1-x^2)}}-x\frac{2x{\sin }^{-1}x}{\sqrt{1-x^2}}-2$

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

$=0$ $\left[henceproved\right]$