Answer to Question #89314 in Differential Equations for lana majeed

Question

solving this ODE by power series x^2y''+(4x-x^2)y'+(2-x)y=0

Answer 1

$$y(x)=\sum_{n=0}^\infin a_nx^n$$

$$y'(x)=\sum_{n=1}^\infin na_nx^{n-1}$$ $$y''(x)=\sum_{n=2}^\infin n(n-1)a_nx^{n-2}$$

$$x^2\sum_{n=2}^\infin n(n-1)a_nx^{n-2}+(4x-x^2)\sum_{n=1}^\infin na_nx^{n-1}+(2-x)\sum_{n=0}^\infin a_nx^n=0$$

$$\sum_{n=2}^\infin n(n-1)a_nx^{n}+\sum_{n=1}^\infin 4na_nx^{n}-\sum_{n=1}^\infin na_nx^{n+1}+\sum_{n=0}^\infin 2a_nx^n-\sum_{n=0}^\infin a_nx^{n+1}=0$$

$$\sum_{n=2}^\infin n(n-1)a_nx^{n}+\sum_{n=1}^\infin 4na_nx^{n}-\sum_{n=2}^\infin (n-1)a_{n-1}x^{n}+\sum_{n=0}^\infin 2a_nx^n-\sum_{n=1}^\infin a_{n-1}x^{n}=0$$

$$2a_0+\sum_{n=1}^\infin [n(n-1)a_n+4na_n- (n-1)a_{n-1}+ 2a_n- a_{n-1}]x^{n}=0$$

$$n=0 ; a_0=0$$

$$a_n=na_{n-1}/[n(n-1)+4n+2]$$

$$y(x)=0$$