Question

If $y=e^{msi{n}^{-1}x}$, then show that $(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}-m^{2}y=0$.

Answer 1

This problem requires the use of chain rule in evaluating derivatives:

$y=e^{m{\sin }^{-1}x}\frac{dy}{dx}=e^{m{\sin }^{-1}x}.\frac{m}{\sqrt{1-x^2}}=\frac{my}{\sqrt{1-x^2}}\frac{d^{2}y}{{dx}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{m}{1-x^2}.(\sqrt{1-x^2}.\frac{dy}{dx}+\frac{xy}{\sqrt{1-x^2}})=\frac{m}{1-x^2}.(\sqrt{1-x^2}.\frac{my}{\sqrt{1-x^2}}+\frac{xy}{\sqrt{1-x^2}})=\frac{m}{1-x^2}.(my+\frac{xy}{\sqrt{1-x^2}})\Rightarrow x\frac{dy}{dx}=\frac{mxy}{\sqrt{1-x^2}},(1-x^2)\frac{d^{2}y}{{dx}^2}=m.(my+\frac{xy}{\sqrt{1-x^2}})=m^{2}y+\frac{mxy}{\sqrt{1-x^2}}\Rightarrow (1-x^2)\frac{d^{2}y}{{dx}^2}=m^{2}y+x\frac{dy}{dx}\Rightarrow (1-x^2)\frac{d^{2}y}{{dx}^2}-x\frac{dy}{dx}-m^{2}y=0$