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If $y=\sin \left(m \sin ^{-1} x\right)$, then $\left(1-x^2\right) y_2-x y_1$ is equal to $\left(\right.$ Here, $y_n$ denotes $\left.\frac{d^n y}{d x^n}\right)$
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The correct answer is:
$-m^2 y$
$y=\sin \left(m \sin ^{-1} x\right)$
$y_1=\cos \left(m \sin ^{-1} x\right) \cdot m \cdot \frac{1}{\sqrt{1-x^2}}$ where $\left(y_1=\frac{d y}{d x}\right)$
$y_1 \sqrt{1-x^2}=m \cos \left(m \sin ^{-1} x\right)$
$\Rightarrow y_2 \sqrt{1-x^2}+y_1 \frac{1}{2 \sqrt{1-x^2}} \cdot(-2 x)$
$=-m \sin \left(m \sin ^{-1} x\right) \cdot \frac{m}{\sqrt{1-x^2}}$
$\left(\because y_2=\frac{d^2 y}{d x^2}\right)$
$\Rightarrow y_2 \sqrt{1-x^2}-\frac{x y_1}{\sqrt{1-x^2}}$
$=\frac{-m^2}{\sqrt{1-x^2}} \sin \left(m \sin ^{-1} x\right)$
$\Rightarrow \quad y_2\left(1-x^2\right)-x y_1=-m^2 y$
$y_1=\cos \left(m \sin ^{-1} x\right) \cdot m \cdot \frac{1}{\sqrt{1-x^2}}$ where $\left(y_1=\frac{d y}{d x}\right)$
$y_1 \sqrt{1-x^2}=m \cos \left(m \sin ^{-1} x\right)$
$\Rightarrow y_2 \sqrt{1-x^2}+y_1 \frac{1}{2 \sqrt{1-x^2}} \cdot(-2 x)$
$=-m \sin \left(m \sin ^{-1} x\right) \cdot \frac{m}{\sqrt{1-x^2}}$
$\left(\because y_2=\frac{d^2 y}{d x^2}\right)$
$\Rightarrow y_2 \sqrt{1-x^2}-\frac{x y_1}{\sqrt{1-x^2}}$
$=\frac{-m^2}{\sqrt{1-x^2}} \sin \left(m \sin ^{-1} x\right)$
$\Rightarrow \quad y_2\left(1-x^2\right)-x y_1=-m^2 y$
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