Q108 / 160 · TS EAMCET 2009 · Mathematics
$y=e^{a \sin ^{-1} x} \Rightarrow\left(1-x^2\right) y_{n+2}-(2 n+1) x y_{n+1}$ is equal to
- A$-\left(n^2+a^2\right) y_n$
- B$\left(n^2-a^2\right) y_n$
- Correct answerC$\left(n^2+a^2\right) y_n$
- D$-\left(n^2-a^2\right) y_n$
Solution
Given, $\quad y=e^{a \sin ^{-1} x}$
On differentiating w.r.t. $x$, we get
$\begin{aligned}
& y_1=e^{a \sin ^{-1} x} a \cdot \frac{1}{\sqrt{1-x^2}} \\
\Rightarrow \quad & y_1 \sqrt{1-x^2}=a y \\
\Rightarrow \quad & \left(1-x^2\right) y_1^2=a^2 y^2
\end{aligned}$
Again, differentiating w.r.t. $x$, we get
$\begin{aligned}
\left(1-x^2\right) 2 y_1 y_2-2 x y_1^2 & =a^2 2 y y_1 \\
\Rightarrow \quad\left(1-x^2\right) y_2-x y_1-a^2 y & =0
\end{aligned}$
Using Leibnitz's rule,
$\begin{aligned}
& \left(1-x^2\right) y_{n+2}+{ }^n C_1 y_{n+1}(-2 x)+{ }^n C_2 y_n(-2) \\
& -x y_{n+1}-{ }^n C_1 y_n-a^2 y_n=0 \\
& \Rightarrow\left(1-x^2\right) y_{n+2}+x y_{n+1}(-2 n-1) \\
& +y_n\left[-n(n-1)-n-a^2\right]=0 \\
& \Rightarrow\left(1-x^2\right) y_{n+2}-(2 n+1) x y_{n+1}=\left(n^2+a^2\right) y_n \\
&
\end{aligned}$
Hence, the correct answer is option: 'C'.