Question

solving xy''+2y'+xy=0 by power series

Answer 1

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

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

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

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

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

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

$$2a_1+\sum_{n=2}^\infin [n(n-1)a_n+2na_n+a_{n-2}]x^{n-1}=0$$

$$a_1=0$$

For

$$n\ge2$$

$$n(n-1)a_n+2na_n+a_{n-2}=0$$

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

So

$$a_2=-a_0/6=-a_0/(2*3)$$

$$a_3=a_5=a_7=a_9=...=0$$

$$a_4=-a_2/20=a_0/120=a_0/(2*3*4*5)$$

$$a_6=-a_4/42=-a_0/(2*3*4*5*6*7)$$

$$y(x)=a_0(1-x^2/(2*3)+x^4/(2*3*4*5)-x^6/(2*3*4*5*6*7)+(-1)^kx^{2k}/(2k+1)!)$$

$$y(x)=a_0 \sum_{k=0}^ \infin(-1)^kx^{2k}/(2k+1)!$$