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 $y = 1 + x + x^2 + x^3 + . . . + \infty$ and $|x| < 1$,then $y^{\prime \prime} =$

If $x = \log p$ and $y = \frac{1}{p}$,then

If $y=ax^{n+1}+b x^{-n}$,then $x^2 \frac{d^2 y}{d x^2}=$

If $f(x) = \sin x + \cos x$, then $f\left(\frac{\pi}{4}\right) f^{(iv)}\left(\frac{\pi}{4}\right)$ is equal to

Suppose, $f(x)=e^{-\sqrt{x}}+e^{-\frac{1}{x^2}}$. If $f^{\prime \prime}(x)=\alpha \cdot \frac{e^{-\sqrt{x}}}{x}\left(1+\frac{1}{\sqrt{x}}\right)+\beta \cdot \frac{e^{-\frac{1}{x^2}}}{x^4}\left(3-\frac{2}{x^2}\right)$, then $(\alpha, \beta)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module