Question

$y=e^{asin-1}x\Rightarrow (1-x^2)y_n+2-(2n+1)x{y}_n+1$ is equal to

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

Answer 1

Answer: (C)

$(n^2+a^2)y_n$

Given,$y=e^{asin-1}x$
On differentiating w.r.t $x$, we get
$y_1=e^{asin-1}xa.\frac{1}{\overline{1-x^2}}$
$\Rightarrow y_1\overline{1-x^2}=ay$
$\Rightarrow 1-x^2{y}_1^2=a^2{y}^2$
Again, differentiating w.r.t $x$, we get
$(1-x^2)2y_1{y}_2-2x{y}_1{}^2=a^{2}2y{y}_1$
$\Rightarrow (1-x^2)y_2-x{y}_1-a^{2}y=0$
Using Leibntiz's rule,
$(1-x^2)y_{n+2}+{}^n{c}_1{y}_{n+1}(-2x)+{}^n{c}_2{y}_2(-2)-x{y}_{n+1}{-}^n{c}_{1}yn-a^{2}yn=0$
$\Rightarrow (1-x^2)y_{n+2}+x{y}_{n+1}(-2n-1)+y_n[-n(n-1)-n-a^2]=0$
$\Rightarrow (1-x^2)y_{n+2}-(2n+1)x{y}_{n+1}=(n^2+a^2)y_n$