Question

If $y=\sin \left(m{\sin }^{-1}x\right)$, then $\left(1-x^2\right)y_2-x{y}_1$ is equal to $($ Here, $y_n$ denotes $\frac{d^{n}y}{d{x}^n})$

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

Answer 1

Answer: (B)

$y=\sin \left(m{\sin }^{-1}x\right)$
$y_1=\cos \left(m{\sin }^{-1}x\right)\cdot m\cdot \frac{1}{\sqrt{1-x^2}}$
where $\left(y_1=\frac{dy}{dx}\right)$
$y_1\sqrt{1-x^2}=m\cos \left(m{\sin }^{-1}x\right)$
$\Rightarrow y_2\sqrt{1-x^2}+y_1\frac{1}{2\sqrt{1-x^2}}\cdot (-2x)$
$=-m\sin \left(m{\sin }^{-1}x\right)\cdot \frac{m}{\sqrt{1-x^2}}$
$\left(\because y_2=\frac{d^{2}y}{d{x}^2}\right)$
$\Rightarrow y_2\sqrt{1-x^2}-\frac{x{y}_1}{\sqrt{1-x^2}}$
$=\frac{-m^2}{\sqrt{1-x^2}}\sin \left(m{\sin }^{-1}x\right)$
$\Rightarrow y_2\left(1-x^2\right)-x{y}_1=-m^{2}y$