Question

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

Answer 1

$\begin{array}{c}y={\sin }^{-1}x\\ \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}\\ \frac{d^{2}y}{d{x}^2}=\frac{-1}{2}(1-x^2{)}^{\frac{-1}{2}-1}(-2x)\\ \frac{d^{2}y}{d{x}^2}=(1-x^2{)}^{\frac{-3}{2}}\left(x\right)\\ \frac{d^{2}y}{d{x}^2}=\frac{x}{1-x^2}\frac{1}{\sqrt{1-x^2}}\\ \frac{d^{2}y}{d{x}^2}=\frac{x}{1-x^2}\frac{dy}{dx}\\ (1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=0\end{array}$