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 $\displaystyle y(x)=\sum_{n=0}^{\infty}c_{n}x^{n}$
The general solution would be,
$\displaystyle y(x)=\sum_{n=0}^{\infty}c_{n}x^{n}=\sum^{\infty}_{ k=0}c_{2k}x^{2k}+\sum_{k=0}^{\infty}c_{2k+1}x^{2k+ 1}$
You have found out that, $\displaystyle c_{2k}=\frac{(-1)^k c_{0}}{(2k)!}$ and $\displaystyle c_{2k+1}=\frac{(-1)^{k}(c_{1}-1)}{(2k+1)!}$ for $\displaystyle k\geq{2}$
(Note: You have written the $\displaystyle c_{2k+1}$ term incorrectly)
Therefore, $\displaystyle y(x)=c_{0}+c_{2}x^{2}+\sum_{k=2}^{\infty}c_{2k}x^{ 2k}+c_{1}x+c_{3}x^{3}+\sum_{k=2}^{\infty}c_{2k+1}x ^{2k+1}$
$\displaystyle y(x)=c_{0}+\left(\frac{-c_{0}}{2}\right)x^{2}+\sum_{k=2}^{\infty}\frac{(-1)^k c_{0}}{(2k)!}x^{2k}+c_{1}x+\left(\frac{-(c_{1}-1)}{6}\right)x^{3}+\sum_{k=2}^{\infty}\frac{(-1)^{k}(c_{1}-1)}{(2k+1)!}x^{2k+1}$
Hope this will help you.