Q74 / 120 · VITEEE 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
$y=e^{a \sin ^{-1} x}$
On differentiating w.r.t. $x$, we get
$$
\begin{array}{l}
y_{1}=e^{a \sin ^{-1} x} a \cdot \frac{1}{\sqrt{1-x^{2}}} \\
\Rightarrow y_{1} \sqrt{1-x^{2}}=a y \\
\Rightarrow\left(1-x^{2}\right) y_{1}^{2}=a^{2} y^{2}
\end{array}
$$
Again differentiating wr.t. $x$, we get
$$
\begin{array}{l}
\left(1-x^{2}\right) 2 y_{1} y_{2}-2 x y_{1}^{2}=a^{2} 2 y y_{1} \\
\Rightarrow\left(1-x^{2}\right) y_{2}-x y_{1}-a^{2} y=0
\end{array}
$$
Using Leibnitz's rule,
$$
\begin{array}{c}
\left(1-x^{2}\right) y_{n+2}+{ }^{n} C_{1} y_{n+1}(-2 x)+{ }^{n} C_{2} y_{n}(-2) \\
-x y_{n+1}-{ }^{n} C_{1} y_{n}-a^{2} y_{n}=0 \\
\Rightarrow\left(1-x^{2}\right) y_{n+2}+x y_{n+1}(-2 n-1) \\
\quad+y_{n}\left[-n(n-1)-n-a^{2}\right]=0 \\
\Rightarrow\left(1-x^{2}\right) y_{n+2}-(2 n+1) x y_{n+1}=\left(n^{2}+a^{2}\right) y_{n}
\end{array}
$$
Hence, the correct answer is option: 'C'.