Question

If y=etan_1x. prove that `(1+x^2)yn+2[2(n+1)x-1]y_n+1+n(n+1)y_n=0`

Answer 1

`y=e^tan-1 x`                     ...................(1) 

Diff w.r.t x, 

`y_1=e^tan-1  x 1/(x^2+1)`

`(x^2+1)y_1= e tan^-1 x=y`         ..................(from 1) 

Again diff. w.r.t x, 

`(x^2+1)y_2+2xy_1=y_1`                  ...............(1)

Now take n th derivative by applying Leibnitz theorem,
Leibnitz theorem is :

`(uv)_n=u_nv+"_1^nCu_n-1v_1+ _2^nCu_(n-2) v_2+....+uv_n` 

`u=(x^2+1),v=y^2`   ................. for first term in eqn(1)

`u=2x,v=y_1`              .............. for second term in eqn (1) 

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

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

Hence Proved.