# y=sin(msin^-1x), then check whether or not (1-x^2)y_(n+2)-(2n+1)xy_(n+1)+(m^2-n^2)y_n=0, also find y_n(0)?

May 10, 2018

#### Explanation:

Here,

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

Diff.w.r.t.$x$

${y}_{1} = \cos \left(m {\sin}^{-} 1 x\right) \frac{d}{\mathrm{dx}} \left(m {\sin}^{-} 1 x\right)$

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

$\sqrt{1 - {x}^{2}} \cdot {y}_{1} = m \cos \left(m {\sin}^{-} 1 x\right)$

Squaring both sides

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

$\left(1 - {x}^{2}\right) {y}_{1}^{2} = {m}^{2} \left(1 - {y}^{2}\right) \ldots \to \left(2\right)$

Again diff.w.r.t.$x$

$\left(1 - {x}^{2}\right) 2 {y}_{1} {y}_{2} - 2 x {y}_{1}^{2} = {m}^{2} \left(- 2 y {y}_{1}\right) \ldots \to \left(3\right)$

Dividing both sides by $2 {y}_{1}$

$\left(1 - {x}^{2}\right) {y}_{2} - x {y}_{1} + {m}^{2} y = 0. . . \to \left(4\right)$

$\text{Using "color(blue)"Leibnitz's Theorem}$ to diff.each term of  (4), n times

(i)D^n(1-x^2)y_2= color(green)(y_(n+2)(1-x^2))+color(red)(ny_(n+1)(-2x))+color(blue)((n*(n-1))/(1*2)y_n(-2)to(A)

(ii)D^n(-xy_1)=color(red)(-y_(n+1)(x)-color(blue)(ny_n(1)...to(B)

(iii)D^n(m^2y)=color(blue)(m^2y_n...to(C)

Adding right hand side terms of $\left(A\right) , \left(B\right) , \mathmr{and} \left(C\right)$

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

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

[contd.]

May 10, 2018

$\mathmr{if} , n$ is even then ${y}_{n} \left(0\right) = 0$

And $n , i s , o \mathrm{dd} \implies \left(n - 2\right)$ is odd

$\implies {y}_{n - 2} = \left[{\left(n - 2\right)}^{2} - {m}^{2}\right] \ldots \left({3}^{2} - {m}^{2}\right) \left({1}^{2} - {m}^{2}\right) \cdot m$

#### Explanation:

we continue the answer from the point reached in Part-1

For ${y}_{n} \left(0\right)$ , we take $x = 0$ in $\left(1\right) , \left(2\right) , \left(3\right) , \left(4\right)$

$\left(1\right) y = \sin \left(m {\sin}^{-} 1 x\right) \implies y = 0$

$\left(2\right) \left(1 - {x}^{2}\right) {y}_{1}^{2} = {m}^{2} \left(1 - {y}^{2}\right) \implies {y}_{1} = m$

$\left(3\right) \left(1 - {x}^{2}\right) {y}_{2} - x {y}_{1} + {m}^{2} y = 0 \implies {y}_{2} = 0$

$\left(4\right) \left(1 - {x}^{2}\right) {y}_{n + 2} - \left(2 n + 1\right) x {y}_{n + 1} + \left({m}^{2} - {n}^{2}\right) {y}_{n} = 0$

$\implies {y}_{n + 2} = - \left({m}^{2} - {n}^{2}\right) {y}_{n} \mathmr{and}$ putting $n = 2$

$\implies {y}_{4} = - \left({m}^{2} - {2}^{2}\right) {y}_{2} = 0. . . \to \left[a s , {y}_{2} = 0\right]$

Similarly,we get, ${y}_{6} = 0 , {y}_{8} = 0 , {y}_{10} = 0 , {y}_{12} = 0. . .$

$i . e . \mathmr{if} , n$ is even then ${y}_{n} \left(0\right) = 0$

Again putting $n = 1 , 3 , 5 , 7 , \ldots$ (n is odd) and $x = 0$ in $\left(4\right)$

$n = 1 \implies {y}_{3} + \left({m}^{2} - {1}^{2}\right) {y}_{1} = 0 \implies {y}_{3} = \left({1}^{2} - {m}^{2}\right) \cdot m$

$n = 3 \implies {y}_{5} + \left({m}^{2} - {3}^{2}\right) {y}_{3} = 0$

$\implies {y}_{5} = \left({3}^{2} - {m}^{2}\right) \left({1}^{2} - {m}^{2}\right) \cdot m$

$n = 5 \implies {y}_{7} = \left({5}^{2} - {m}^{2}\right) \left({3}^{2} - {m}^{2}\right) \left({1}^{2} - {m}^{2}\right) \cdot m$
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And $n , i s , o \mathrm{dd} \implies \left(n - 2\right)$ is odd

$\implies {y}_{n - 2} = \left[{\left(n - 2\right)}^{2} - {m}^{2}\right] \ldots \left({3}^{2} - {m}^{2}\right) \left({1}^{2} - {m}^{2}\right) \cdot m$