y=e^{a sin ^{-1} x} Rightarrow (1-x^2) y_{n+2 | vedclass

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(C) Given,$y=e^{a \sin ^{-1} x}$.On differentiating with respect to $x$,we get:$y_1 = e^{a \sin ^{-1} x} \cdot \frac{a}{\sqrt{1-x^2}}$$\Rightarrow y_1

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$\left(n^2+a^2\right) y_n$

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(C) Given,$y=e^{a \sin ^{-1} x}$.On differentiating with respect to $x$,we get:$y_1 = e^{a \sin ^{-1} x} \cdot \frac{a}{\sqrt{1-x^2}}$$\Rightarrow y_1 \sqrt{1-x^2} = ay$Squaring both sides:$(1-x^2) y_1^2 = a^2 y^2$Differentiating again with respect to $x$:$(1-x^2) 2 y_1 y_2 - 2x y_1^2 = a^2 2y y_1$Dividing by $2y_1$ (assuming $y_1 
eq 0$):$(1-x^2) y_2 - x y_1 - a^2 y = 0$Applying Leibnitz's theorem for the $n$-th derivative:$[(1-x^2) y_{n+2} + n(-2x) y_{n+1} + \frac{n(n-1)}{2}(-2) y_n] - [x y_{n+1} + n(1) y_n] - a^2 y_n = 0$$(1-x^2) y_{n+2} - 2nx y_{n+1} - n(n-1) y_n - x y_{n+1} - n y_n - a^2 y_n = 0$$(1-x^2) y_{n+2} - (2n+1) x y_{n+1} - (n^2 - n + n + a^2) y_n = 0$$(1-x^2) y_{n+2} - (2n+1) x y_{n+1} = (n^2 + a^2) y_n$

$y=e^{a \sin ^{-1} x} \Rightarrow (1-x^2) y_{n+2}-(2 n+1) x y_{n+1}$ is equal to

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