Hi,

Can you please check this for me?

Thanks,

James

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

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- May 12th 2011, 03:39 PMjames121515Power series method problem
Hi,

Can you please check this for me?

Thanks,

James

http://i708.photobucket.com/albums/w...5/powerser.jpg - May 12th 2011, 05:52 PMSudharaka
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.