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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Aicharya · Accepted Answer

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

If $y=\sin \left(m \sin ^{-1} x\right)$, then $\left(1-x^2\right) y_2-x y_1$ is equal to $\left(\right.$…

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