Question

$y=e^{a{\sin }^{-1}x}$, then $\left(1-x^2\right)y_{n+2}-\left(2n+1\right)x{y}_{n+1}$ is equal to

• A. $-\left(n^2+a^2\right)y_n$
• B. $\left(n^2-a^2\right)y_n$
• C. $\left(n^2+a^2\right)y_n$
• D. $-\left(n^2-a^2\right)y_n$

Answer 1

The correct option is C

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

Explanation for the correct option:

Step $1$: Finding the first and second derivatives

The given equation is,

$y=e^{a{\sin }^{-1}x}$

Differentiate both sides with respect to $x$, we get

$y_1=e^{a{\sin }^{-1}x}a\frac{1}{\sqrt{1-x^2}}$

$\Rightarrow$ $y_1\sqrt{1-x^2}=a{e}^{a{\sin }^{-1}x}$

Squaring both sides, we get

${y_1}^2\left(1-x^2\right)=a^2{y}^2$

again differentiate both sides with respect to $x$, we get

$2y_1{y}_2\left(1-x^2\right)+{y_1}^2\left(-2x\right)=2y{a}^2{y}_1$

$\Rightarrow$ $2y_1\left[y_2\left(1-x^2\right)-x{y}_1-y{a}^2\right]=0$

$\Rightarrow$ $y_1\left[y_2\left(1-x^2\right)-x{y}_1-y{a}^2\right]=0$

Step $2$: Finding the required equation

Leibnitz theorem for the $n^{th}$ derivative of a product:

$\frac{d^n}{d{x}^n}\left(f\left(x\right)g\left(x\right)\right)=f^{\left(n\right)}\left(x\right)g\left(x\right)+\frac{n}{1!}{f}^{\left(n-1\right)}\left(x\right)g\text{'}\left(x\right)+\frac{n\left(n-1\right)}{2!}{f}^{\left(n-2\right)}\left(x\right)g^{\left(2\right)}\left(x\right)+....+f\left(x\right)g^{\left(n\right)}\left(x\right)$

Using the Leibnitz theorem, we get

$y_{n+2}\left(1-x^2\right)-\left(2n+1\right)x{y}_{n+1}-y_n\left(n^2+a^2\right)=0$

$\Rightarrow$ $y_{n+2}\left(1-x^2\right)-\left(2n+1\right)x{y}_{n+1}=y_n\left(n^2+a^2\right)$

Hence, the correct option is $C)\left(n^2+a^2\right)y_n$.