Question

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

Answer 1

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

Differentiating w.r.t. $x$,

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

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

Squaring both sides $(1-x^2){\left(\frac{dy}{dx}\right)}^2=4({\sin }^{-1}x{)}^2$

$\Rightarrow (1-x^2){\left(\frac{dy}{dx}\right)}^2=4y$

Again differentiating w.r.t. x

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

Dividing both sides by $2\frac{dy}{dx}$

$(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=2$. $\left[henceproved\right]$