Question

$y=e^{m{\sin }^{-1}x}$, then $(1-x^2)\frac{d^{2}y}{d{x}^x}-x\frac{dy}{dx}=$

• A. $m^2$
• B. $m^{2}y$
• C. $2m^{2}y$
• D. $\frac{m^2}{2}y$

Answer 1

The correct option is B $m^{2}y$

$y=e^{msi{n}^{1}x}$

$y^{\prime }=\frac{m{e}^{msi{n}^{-1}x}}{\sqrt{1-x^2}}$

$y^{\prime \prime }=\left[\frac{\left(\frac{m^2{e}^{msi{n}^{-1}x}}{\sqrt{1-x^2}-\left(m^{emsi{n}^1}\right)\frac{(-2x)}{2\sqrt{1-x^2}}}\right)}{(\sqrt{1-x^2}{)}^2}\right]$

$=\frac{m^2{e}^{msi{n}^{-1}x}+\frac{mx{e}^{msi{n}^{-1}x}}{\sqrt{1-x^2}}}{(1-x^2)}$

$=\frac{m^2{e}^{si{n}^{-1}x+\frac{mx{e}^{msi{n}^{1}x}}{\sqrt{1-x^2}}}}{(1-x^2)}$

$=\frac{m^{2}y+x{y}^1}{(1-x^2)}$

$(1-x^2)y^{11}=m^{2}y+x{y}^1$

$(1-x^2)y^{11}-x{y}^1=m^{2}y$

$(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=m^{2}y$

So, option (B) correct.