Answer to Question #128508 in Differential Equations for Zubair

Question

Solve the Lagrange's equation (1-x^2)y"-2xy'+n(n+1)y=0

Answer 1

$Let \quad y(x)=\sum\limits_{k=0}^{\infty}a_kx^k \quad then \\ (1-x^2)y''-2xy'+n(n+1)y=2(n(n+1)a_0+a_2)+\\ +2x(6a_3+2a_1(3n(n+1)-1))+\\ +\sum\limits_{k=2}^{\infty}((k(k-3)+n(n+1))a_k-(k+1)(k+2)a_{k+2})x^k=0\Rightarrow\\ a_2=-n(n+1)a_0,\quad a_3=a_1(\frac{1}{3}-n(n+1)),\\ a_{k+2}=\frac{((k(k-3)+n(n+1))}{(k+1)(k+2)}a_k.\\ result:\quad y_{general}=y(a_0,a_1,x)=\sum\limits_{k=0}^{\infty}a_kx^k\quad and\\ a_2=-n(n+1)a_0,\quad a_3=a_1(\frac{1}{3}-n(n+1)),\\ a_{k+2}=\frac{((k(k-3)+n(n+1))}{(k+1)(k+2)}a_k;$