Here, #y_1,y_2,...,y_n# denotes the forst, second,...,#n^(th)# derivative of #y#.
Differentiating, using the Chain Rule, the given eqn. w.r.to #x,#
#dy/dx=y_1=1/{x+sqrt(x^2+1)}d/dx{x+sqrt(x^2+1)}#,
#=1/{x+sqrt(x^2+1)}{d/dx x+d/dx sqrt(x^2+1)}#,
#=1/{x+sqrt(x^2+1)}{1+1/(2sqrt(x^2+1))d/dx(x^2+1)}#,
#=1/{x+sqrt(x^2+1)}{1+1/(2sqrt(x^2+1))(2x)}#,
#=1/{x+sqrt(x^2+1)}{1+x/sqrt(x^2+1)}#,
#=1/{x+sqrt(x^2+1)}{sqrt(x^2+1)+x}/sqrt(x^2+1)#.
#rArr y_1=1/sqrt(x^2+1)#.
#rArr y_1sqrt(x^2+1)=1#.
Diff.ing again, using the Product Rule, w.r.to #x,# we get,
#y_1d/dxsqrt(x^2+1)+sqrt(x^2+1)d/dxy_1=0#,
#:. y_1{(1/(2sqrt(x^2+1)))(2x)}+sqrt(x^2+1)y_2=0, or#
#(xy_1)/sqrt(x^2+1)+y_2sqrt(x^2+1)=0, i.e.,#
#(x^2+1)y_2+xy_1=0.#
#:. (y_2(x^2+1))_n+(y_1x)_n=0...............(star)#.
Now, we use the following Leibnitz Rule :
#(uv)_n=u_nv+""_nC_1u_(n-1)v_1+""_nC_2u_(n-2)v_2+...+v_n#.
We take #u=y_2 and v=x^2+1#.
#:. (y_2(x^2+1))_n#
#=y_((n+2))(x^2+1)+ny_(n+1)(2x)+1/2*n(n-1)y_n*2+0#,
#=(x^2+1)y_(n+2)+2nxy_(n+1)+n(n-1)y_n..........(star_1)#.
Similarly, #(y_1x)_n,#
#=y_(n+1)x+ny_n..................(star_2)#.
Then, #(star),(star_1) and (star_2),#
#rArr (x^2+1)y_(n+2)+2nxy_(n+1)+n(n-1)y_n+y_(n+1)x+ny_n=0#.
#rArr (x^2+1)y_(n+2)+(2n+1)xy_(n+1)+{n(n-1)+n}y_n=0#.
#rArr (x^2+1)y_(n+2)+(2n+1)xy_(n+1)+n^2y_n=0.#
Q.E.D.
