Question

If y= cos (msin_1 x).Prove that `(1-x^2)y_n+2-(2n+1)xy_(n+1)+(m^2-n^2)y_n=0`

Answer 1

`y = cos(msin^-1x)`

Differentiating w.r.t. ‘x’, y1 = `-sin (msin_1x).m.1/sqrt(1-x^2)`

∴ `sqrt(1-x^2.y_1)=- msin(msin^-1x)`

On Squaring, `(1-x^2)y_1^2=m^2 sin^2 (msin^-1x)`

∴ `(1-x)^2 y_1^2=m^2[1-cos^2(msin_1x)]`

∴ `(1-x^2)y_1^2=m^2[1-y^2]`              (From 1) 

Again differentiating w.r.t. 'x',

`(1-x^2)2y_1y_2+y_1^2(-2x)=m^2(0-2yy_1)`

∴ `(1-x^2)y_1-xy_1=-m^2y`              (Dividing by 2y1) 

Applying Leibnitz theorem,` {y_n=u_nv+nu_(n-1)v_1+^nC_2u_(n-2)v_2+^nC_3u_(n-3)v_3+........}`

`[(1-x^2)y_(n+2)+n(-2x)y_(n+1)+n(n-1)/(2!)]-[xy_n+1+ny_n]=-m^2y_n`

∴` (1-x^2)y_(n+2)+n(-2x)y_n+1+(n(n-1))/(2!)-xy_(n+1)-ny_n+m^2y_n=0`

∴` (1-x^2)y_n+2-xy_n(2n+1)+(-n^2+n-n+m^2)y_n=0`

∴ `(1-x^2)y_(n+2)-(2n+1)xy_n+1+(m^2-n^2)y_n=0`