Q127 / 200 · Manipal (MET) 2010 · Mathematics
If $y=\sin \left(m \sin ^{-1} x\right)$, then $\left(1-x^2\right) y^{\prime \prime}-x y^{\prime}$ is equal to
- A$m^2 y$
- B$m y$
- Correct answerC$-m^2 y$
- DNone of these
Solution
We have, $y=\sin \left(m \sin ^{-1} x\right)$
$\frac{d y}{d x}=\cos \left(m \sin ^{-1} x\right) \cdot \frac{m}{\sqrt{1-x^2}}$
$\begin{aligned} \Rightarrow\left(1-x^2\right)\left(\frac{d y}{d x}\right)^2 & =m^2 \cos ^2\left(m \sin ^{-1} x\right) \\ & =m^2(1- )\end{aligned}$
On differentiating w.r.t. $x$, again, we get $\left(1-x^2\right) 2 \cdot \frac{d y}{d x} \cdot \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2 \cdot(-2 x)$
$=-2 m^2 y \frac{d y}{d x}$
$\Rightarrow \left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}=-m^2 y$
$[\because$ Dividing by $2 d y / d x]$
Hence, the correct answer is option: 'C'.