Question

If $y=e^{m{\sin }^{-1}x},-1\le x\le 1$, show that $(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}-a^{2}y=0$

Answer 1

$y=e^{aco{s}^{-1}x}\frac{dy}{dx}=e^{a{\cos }^{-1}x}\times a\frac{-1}{\sqrt{1-x^2}}\frac{d^{2}y}{d{x}^2}=e^{a{\cos }^{-1}x}\times \frac{a^2}{1-x^2}-e^{aco{s}^{-1}x}\times a\times \frac{1}{2}\times \frac{2x}{(1-x^2{)}^{\frac{3}{2}}}$
Now,
$(1-x^2{)}^2\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}-a^{2}y{a}^2{e}^{a{\cos }^{-1}x}-e^{a{\cos }^{-1}x}\times a\times \frac{x}{(1-x^2{)}^{\frac{1}{2}}}+x\times (e^{a{\cos }^{-1}x}\times a\frac{-1}{\sqrt{1-x^2}})-a^2\times e^{a{\cos }^{-1}x}=0$