Question

If $y=e^{m{\sin }^{-1}x}$, then $(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}-ky=0$, where $k$ is equal to

• A. $m^2$
• B. $2$
• C. $-1$
• D. $-m^2$

Answer 1

The correct option is A $m^2$

$y=e^{m{\sin }^{-1}x},\frac{dy}{dx}=\frac{m{e}^{m{\sin }^{-1}x}}{\sqrt{1-x^2}}=\frac{my}{\sqrt{1-x^2}}\frac{d}{dx}\cdot \frac{dy}{dx}=m\frac{\sqrt{1-x^2}\cdot \frac{m{e}^{m{\sin }^{-1}x}}{\sqrt{1-x^2}}+e^{m{\sin }^{-1}x}\frac{x}{\sqrt{1-x^2}}}{1-x^2}(1-x^2)\frac{d^{2}y}{d{x}^2}=m^{2}y+x\frac{dy}{dx}(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}-m^{2}y=0$
Comparing this equation with the given equation we have
$k=m^2$