Question #8826d

1 Answer
Jan 12, 2018

Answer:

# y_(n+2)+(2nx+1)y_(n+1)+n(n+1)y_n=0#.

Explanation:

#y=tan^-1x#.

#:. dy/dx=y_1=1/(1+x^2)#.

#:. (1+x^2)y_1=1#.

Re-differentiating w.r.t. #x,# using the Product Rule,

#(1+x^2)(y_1)'+y_1(1+x^2)'=0#.

#:. (1+x^2)y_2+2xy_1=0#.

#:.((1+x^2)y_2)_n+(2xy_1)_n=0.............................(star)#.

Now, we use the following Leibnitz Rule (L) for the #n^(th)#

derivative of product of functions :

#(uv)_n=u_n+""_nC_1u_(n-1)v_1+""_nC_2u_(n-2)v_2+...+v_n#.

Applying L to #(1+x^2)y_2," with, "u=y_2, and v=1+x^2#,

#((1+x^2)y_2)_n,#

#=(y_2)_n+n*(y_2)_(n-1)(2x)+(n(n-1))/2*(y_2)_(n-2)(2)+0#,

#=y_(n+2)+2nxy_(n+1)+n(n-1)y_n......(star^1)#.

Similarly, #(2xy_1)_n,#

#=y_(n+1)+n*y_n*2+0..............................(star^2)#.

#(star), (star^1), (star^2)rArr y_(n+2)+(2nx+1)y_(n+1)+n(n+1)y_n=0#.