Question

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

Answer 1

$\text{Given,}y=({\sin }^{-1}x{)}^2$
Differentiating with respect to $x$, we get
$\frac{dy}{dx}=2{\sin }^{-1}x\times \frac{1}{\sqrt{1-x^2}}$
$\Rightarrow \sqrt{1-x^2}\frac{dy}{dx}=2{\sin }^{-1}x$
Again, differentiating with respect to $x,$ we get
$\sqrt{1-x^2}\frac{d^{2}y}{d{x}^2}-\frac{dy}{dx}\times \frac{2x}{2\sqrt{1-x^2}}=2\times \frac{1}{\sqrt{1-x^2}}$
$\Rightarrow (1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=2$
$\Rightarrow (1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}-2=0$