Question

`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)`?

Answer 1

Please see below.

Explanation:

Here,

`y=sin(msin^-1x)...to(1)`

Diff.w.r.t.`x`

`y_1=cos(msin^-1x)d/(dx)(msin^-1x)`

`y_1=cos(msin^-1x)xxm/sqrt(1-x^2)`

`sqrt(1-x^2)*y_1=mcos(msin^-1x)`

Squaring both sides

`(1-x^2)y_1^2=m^2cos^2(msin^-1x)=m^2(1-sin^2(msin^-1x))`

`(1-x^2)y_1^2=m^2(1-y^2)...to(2)`

Again diff.w.r.t.`x`

`(1-x^2)2y_1y_2-2xy_1^2=m^2(-2yy_1)...to(3)`

Dividing both sides by `2y_1`

`(1-x^2)y_2-xy_1+m^2y=0...to(4)`

`"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 `(A),(B),and(C)`

`y_(n+2)(1-x^2)+y_(n+1)[-2nx-x]+y_n[-n^2+n-n+m^2]=0`

`=>(1-x^2)y_(n+2)-(2n+1)xy_(n+1)+(m^2-n^2)y_n=0`

Please see the second answer.
[contd.]

Answer 2

`if,n` is even then `y_n(0)=0`

And `n ,is ,odd=>(n-2)` is odd

`=>y_(n-2)=[(n-2)^2-m^2]...(3^2-m^2)(1^2-m^2)*m`

Explanation:

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

For `y_n(0)` , we take `x=0` in `(1),(2),(3),(4)`

`(1)y=sin(msin^-1x)=>y=0`

`(2)(1-x^2)y_1^2=m^2(1-y^2)=>y_1=m`

`(3)(1-x^2)y_2-xy_1+m^2y=0=>y_2=0`

`(4)(1-x^2)y_(n+2)-(2n+1)xy_(n+1)+(m^2-n^2)y_n=0`

`=>y_(n+2)=-(m^2-n^2)y_n and` putting `n=2`

`=>y_4=-(m^2-2^2)y_2=0...to[as,y_2=0]`

Similarly,we get, `y_6=0,y_8=0,y_10=0,y_12=0...`

`i.e.if,n` is even then `y_n(0)=0`

Again putting `n=1,3,5,7,...` (n is odd) and `x=0` in `(4)`

`n=1=>y_3+(m^2-1^2)y_1=0=>y_3=(1^2-m^2)*m`

`n=3=>y_5+(m^2-3^2)y_3=0`

`=>y_5=(3^2-m^2)(1^2-m^2)*m`

`n=5=>y_7=(5^2-m^2)(3^2-m^2)(1^2-m^2)*m`
...........................

And `n ,is ,odd=>(n-2)` is odd

`=>y_(n-2)=[(n-2)^2-m^2]...(3^2-m^2)(1^2-m^2)*m`