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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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

Similar Questions

If $f(x) = f'(x) + f''(x) + f'''(x) + \ldots$ and $f(0) = 1$,then $f(x)$ is equal to

If $f(x)=\tan ^{-1}\left[\frac{\log \left(\frac{e}{x^{2}}\right)}{\log \left(e x^{2}\right)}\right]+\tan ^{-1}\left[\frac{3+2 \log x}{1-6 \log x}\right]$ then the value of $f^{\prime \prime}(x)$ is equal to

If $x = A \cos 4t + B \sin 4t$,then $\frac{d^2x}{dt^2} = $

If $y = a^x \cdot b^{2x-1}$,then $\frac{d^2 y}{d x^2}$ is equal to

If $x^2+y^2+\sin y=4$,then the value of $\frac{d^2 y}{d x^2}$ at $x=-2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module