Question

If $y=e^{msi{n}^{-1}x}$ then show that $(1-x^2)y_{n+2}-(2n+1)x{y}_{n+1}-(n^2+m^2)y_n=0$.

Answer 1

Question has a problem.

It should be $(1-x^2)y_{n+2}-(2n+1)x{y}_{n+1}-(n^2+m^2)y_n=0$

It should be $y=e^{m{\sin }^{-1}x}$

Differentiating wrt $x$

$y_1=e^{m{\sin }^{-1}x}\left(\frac{m}{{(1-x^2)}^{\frac{1}{2}}}\right)$

$\Rightarrow y_1{(1-x^2)}^{\frac{1}{2}}=my$

$\Rightarrow {y_1}^2(1-x^2)=m^2{y}^2$

Now, again differentiating wrt $x$,

$2y_1(1-x^2)y_2-\left(2x\right){y_1}^2=2m^{2}y{y}_1$

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

Now, general Lebniz rule is that,

${\left(fg\right)}^n\left(x\right)={\sum }_{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)f^{(n-k)}\left(x\right)g^{\left(k\right)}\left(x\right)$

$\left(\begin{array}{c}n\\ k\end{array}\right)=\frac{n!}{k!(n-k)!}{=}^n{C}_k$

So, by differentiating equation $\left(1\right)$, $n$ times and applying Leibniz rule.

$y_{n+2}(1-x^2){-}^n{C}_1{y}_{n+1}(-2x){+}^n{C}_2{y}_n(-2)-\left\{y_{n+1}x{+}^n{C}_1{y}_n\right\}-m^2{y}_n=0$

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

$\therefore (1-x^2)y_{n+2}-(2n+1)x{y}_{n+1}-(n^2+m^2)y_n=0$