Question

If $y=\sin \left(m{\sin }^{-1}x\right)\forall x\in (-1,1)$, then $\frac{y_{n+2}\left(0\right)}{y_n\left(0\right)}=$

• A. $m^2+n^2$
• B. $m^2-n^2$
• C. $n^2-m^2$
• D. $-(m^2+n^2)$

Answer 1

Answer: (C)

$n^2-m^2$

$y=\sin \left(m{\sin }^{-1}x\right)$

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

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

$(1-x^2)y_1^2=(1-y^2)m^2$

differentiate on both sides w.r.t x

$(1-x^2)2y_1{y}_2-2x{y}_1^2=-2y{y}_1{m}^2$

$(1-x^2)y_2-2x{y}_1+y{m}^2=0$

$D^n\left[\right(1-x^2\left)y_2\right]=(1-x^2)y_{n+2}+n_{C_1}(-2x)y_{n+1}+n_{C_2}(-2)y_n$

$=(1-x^2)y_{n+2}-2nx{y}_{n+1}-n(n-1)y_n$

$D^n(-x{y}_1)=(-x)y_{n+1}-n{y}_n$

$D^n\left(y{m}^2\right)=y_n{m}^2$

$D^n\left[\right(1-x^2)y_2-x{y}_1+y{m}^2]=(1-x^2)y_{n+2}-2nx{y}_{n+1}-n(n-1)y_n-x\left(y_{n+1}\right)-n{y}_n+y_n{m}^2=0$

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

At $x=0,$

$(1-0^2)y_{n+2}\left(0\right)-(2n+1)\left(0\right)y_1\left(0\right)+(m^2-n^2)y_n\left(0\right)=0$

$y_{n+2}\left(0\right)+(m^2-n^2)y_n\left(0\right)=0$

$\therefore \frac{y_{n+2}\left(0\right)}{y_n\left(0\right)}=-(m^2-n^2)=n^2-m^2$