Question

If $m{\sin }^{-1}x={\log }_{e}y,$ then $(1-x^2)y"-x{y}^{\prime }$ =

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

Answer 1

Answer: (A)

Given, m $si{n}^{-1}x=lo{g}_{e}y$
$\Rightarrow y=e^{msi{n}^{-1}x}$
On differentiating w.r.t. x, we get
$y^{\prime }=e^{msi{n}^{-1}x}X\frac{m}{\sqrt{1-x^2}}$
$\Rightarrow \sqrt{1-x^2}\cdot y^{\prime }=my$
On squaring both sides, we get
$\left(1-x^2\right)y^{\prime 2}=m^2{y}^2$
Again differentiating, we get
$\left(1-x^2\right)2y^{\prime }X{y}^{\prime \prime }+y^{\prime 2}\left(-2x\right)=m^2\cdot 2y{y}^{\prime }$
$\therefore \left(1-x^2\right)y^{\prime \prime }-x{y}^{\prime }=m^{2}y$