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

Answer: (A)

$m^2$

Given : $y=e^{m{\sin }^{-1}x}$

$⟹\frac{dy}{dx}=e^{m{\sin }^{-1}x}\cdot \frac{m}{\sqrt{1-x^2}}$

$⟹\frac{dy}{dx}=\frac{m{e}^{m{\sin }^{-1}x}}{\sqrt{1-x^2}}$ .... $\left(i\right)$

$\frac{d^{2}y}{d{x}^2}=\frac{m{e}^{m{\sin }^{-1}x}}{\sqrt{1-x^2}}\times \frac{m}{\sqrt{1-x^2}}+m{e}^{m{\sin }^{-1}x}\times \left(\frac{-1}{2}\right)\cdot \left(\frac{-2x}{(1-x^2{)}^{\frac{3}{2}}}\right)$

$\therefore \frac{d^{2}y}{d{x}^2}=\frac{m^2{e}^{m{\sin }^{-1}x}}{(1-x^2)}+\left(\frac{mx{e}^{m{\sin }^{-1}x}}{(1-x^2)\sqrt{1-x^2}}\right)$

Now, $(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}-ky=0$

$⟹(1-x^2)[\frac{m^2{e}^{m{\sin }^{-1}x}}{(1-x^2)}+\left(\frac{mx{e}^{m{\sin }^{-1}x}}{(1-x^2)\sqrt{1-x^2}}\right)]-\frac{xm{e}^{m{\sin }^{-1}x}}{\sqrt{1-x^2}}-k{e}^{m{\sin }^{-1}x}=0$

$⟹m^2{e}^{m{\sin }^{-1}x}+\frac{mx{e}^{m{\sin }^{-1}x}}{\sqrt{1-x^2}}-\frac{xm{e}^{m{\sin }^{-1}x}}{\sqrt{1-x^2}}-k{e}^{m{\sin }^{-1}x}=0$

$⟹m^2{e}^{m{\sin }^{-1}x}-k{e}^{m{\sin }^{-1}x}=0$

$⟹m^2-k=0$

$\therefore k=m^2$