Question

If y=(x+√x²-1 ,Prove that

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

Answer 1

`y+(x+sqrtx^2-1)^m`

taking + sign before the radical

`thereforey_1=m[(x+sqrtx^2-1)^(m-1)].[1+x/(sqrtx^2-1)]`

`=m(x+sqrtx^2-1)^m.x/(sqrtx^2-1)=(my)/(sqrtx^2-1)`

`sqrtx^2-1.y_1=my`

Differentiating again w.r.t x,

`sqrtx^2-1.y_2+x/(sqrtx^2-1)y_1=my_1`

`(x^2-1)y_2+xy_1=msqrtx^2-1.y_1=m.   my=my^2`

`(x^2-1)y_2+xy_1-my^2=0`

Hence after applying lebnitz’s theorem we get,

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