y=e^ a ⁻¹ x (1-x^2 ) y_ n+2 -(2 n+1) x y_ n+1 is equal to

Step-by-step solution

Given, y=e^ a ⁻¹ x On differentiating w.r.t. x , we get aligned & y₁=e^ a ⁻¹ x a 1 1-x^2 & y₁ 1-x^2 =a y & (1-x^2 ) y₁^2=a^2 y^2 aligned Again, differentiating w.r.t. x , we get aligned (1-x^2 ) 2 y₁ y₂-2 x y₁^2 & =a^2 2 y y₁ (1-x^2 ) y₂-x y₁-a^2 y & =0 aligned Using Leibnitz's rule, aligned & (1-x^2 ) y_ n+2 + ^n C₁ y_ n+1 (-2 x)+ ^n C₂ y_n(-2) & -x y_ n+1 - ^n C₁ y_n-a^2 y_n=0 & (1-x^2 ) y_ n+2 +x y_ n+1 (-2 n-1) & +y_n [-n(n-1)-n-a^2 ]=0 & (1-x^2 ) y_ n+2 -(2 n+1) x y_ n+1 = (n^2+a^2 ) y_n & aligned