Power series method problem

Quote:

Originally Posted by james121515

Hi,

Can you please check this for me?

Thanks,
James

http://i708.photobucket.com/albums/w...5/powerser.jpg

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

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

Hope this will help you.