# Thread: Power series method problem

1. ## Power series method problem

Hi,

Can you please check this for me?

Thanks,
James

2. Originally Posted by james121515
Hi,

Can you please check this for me?

Thanks,
James

Dear James,

You have done most of it correctly, except in writing the final answer. Since you have taken the trial solution to be $y(x)=\sum_{n=0}^{\infty}c_{n}x^{n}$

The general solution would be,

$y(x)=\sum_{n=0}^{\infty}c_{n}x^{n}=\sum^{\infty}_{ k=0}c_{2k}x^{2k}+\sum_{k=0}^{\infty}c_{2k+1}x^{2k+ 1}$

You have found out that, $c_{2k}=\frac{(-1)^k c_{0}}{(2k)!}$ and $c_{2k+1}=\frac{(-1)^{k}(c_{1}-1)}{(2k+1)!}$ for $k\geq{2}$

(Note: You have written the $c_{2k+1}$ term incorrectly)

Therefore, $y(x)=c_{0}+c_{2}x^{2}+\sum_{k=2}^{\infty}c_{2k}x^{ 2k}+c_{1}x+c_{3}x^{3}+\sum_{k=2}^{\infty}c_{2k+1}x ^{2k+1}$

$y(x)=c_{0}+\left(\frac{-c_{0}}{2}\right)x^{2}+\sum_{k=2}^{\infty}\frac{(-1)^k c_{0}}{(2k)!}x^{2k}+c_{1}x+\left(\frac{-(c_{1}-1)}{6}\right)x^{3}+\sum_{k=2}^{\infty}\frac{(-1)^{k}(c_{1}-1)}{(2k+1)!}x^{2k+1}$