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I have been having a few questions about series solutions to ODE and I found an example that can illustrate my question.

It is just a simple example, say we consider the ODE

$$ y''-xy'-y=0$$ around the ordinary point $x_o=0$

So we know a solution exists of the form $$y= \sum_{n=0}^{\infty}a_nx^{n}$$ and it follows then, $$y'= \sum_{n=1}^{\infty}na_nx^{n-1}$$

and $$y''= \sum_{n=0}^{\infty}(n-1)(n)a_nx^{n-2}$$

So, in this example substituting in we obtain,

$$\sum_{n=2}^{\infty}(n-1)(n)a_nx^{n-2}-\sum_{n=1}^{\infty}(n)a_nx^{n}-\sum_{n=0}^{\infty}a_nx^{n}$$

Now here is the part I get confused. What is the best way to proceed? How can I know, is there any sort of pattern or way to know this?

I know that I have to get all the coefficients of the x to be the same and also have the starting index to be the same. But I also know I need to make sure I am able to see some recurrence relation.

For example I could make $$y''=\sum_{n=0}^{\infty}(n+1)(n+2)a_{n+2}x^{n}$$

Which would make everything to the n, but then I have another non zero index, and it seems like I can't ever figure out how to line it all up so they match!

I hope someone can help me with this,

thanks a lot

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  • $\begingroup$ Do the same thing to $y'$ as you did to $y''$. Then you'll be set. $\endgroup$ – Avi Steiner May 24 '15 at 20:55
  • $\begingroup$ @AviSteiner Thanks for the comment. But because the y' is multiplied by an x, wouldn't this now make it not to the n power? $\endgroup$ – Quality May 24 '15 at 20:57
  • $\begingroup$ Shoot. Forgot. Just start the indices at one, and do the constant coefficient separately. $\endgroup$ – Avi Steiner May 24 '15 at 21:14

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