I have $$y''-xy'-y=0$$ and I'm trying to find the series solution around the ordinary point $x_0=1$. My last post I muscled through to the solution when the ordinary point was $x_0=0$, but this is proving to be tougher. Now I have obtained through power series analysis $$y''-xy'-y=0=\sum_{k=0}^{\infty}[(k+2)(k+1)a_{k+2}-(k+1)a_{k+1}-(k+1)a_k](x-1)^k$$ which yields the recurrence relation $$a_{k+2}=\frac{a_{k+1}+a_{k}}{k+2}$$ WHen I start plugging in sequential "k" values I'm not finding a very good pattern emerging. $$a_2=\frac{a_0}{2!}+\frac{a_1}{2!}$$ $$a_3=\frac{a_0}{3!}+\frac{3a_1}{3!}$$ $$a_4=\frac{4a_0}{4!}+\frac{6a_1}{4!}$$ $$a_5=\frac{8a_0}{5!}+\frac{18a_1}{5!}$$ $$a_6=\frac{28a_0}{6!}+\frac{48a_1}{6!}$$ $$a_7=\frac{76a_0}{7!}+\frac{156a_1}{7!}$$ Outside just writing out term by term, substituting in the appropriate $a_k$, does this have a nice closed form? Or is is just the case that I have the answer with the $a_k$'s that I have

  • $\begingroup$ Generally speaking, what you have already done (getting a recursive form for the coefficients) is the best you can do when finding power series solutions. $\endgroup$ – JohnD Nov 29 '13 at 4:44
  • $\begingroup$ I checked your recurrence relation and it is correct. $\endgroup$ – Amzoti Nov 29 '13 at 5:19

I totally agree with JohnD' answer to your question. You have done the best you can achieve using power series. To try to clarify, let me go to the solution of your ODE which is
$$y=e^{\frac{x^2}{2}} \left(C_2+C_1\sqrt{\frac{\pi }{2}} \text{erf}\left(\frac{x}{\sqrt{2}}\right)\right)$$

As you see, you have two arbitrary constants and then any polynomial expansion of the solution will depend on two constants : these are your $a_0$ and $a_1$ terms.

To me the nicest closed form is the recurrence relation you established; I suggest you do not go further.


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