# Method for solving ODE with power series

when trying to solve second order linear homogeneous variable coefficient ODEs using a power series method, there seem to be two different general forms cropping up in my notes. The first uses an ordinary point $$x_0$$ $$y = \sum_{m = 0}^{\infty}a_m(x-x_0)^m$$ The second uses a factor of $$x^r$$ $$y = x^r\sum_{m = 0}^{\infty}a_mx^m$$

How do I know which of these forms to use for my solution?

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The second is always more general, i.e. if $r=0$ then you'll just recover the first ansatz. – Pedro Tamaroff Jun 22 '13 at 22:48
but if I have limited time (like in an exam) how can I tell if I can use the first one because that would save time? – Sam Jun 22 '13 at 23:15
The second solution can also be "centered" at $x_o$ by replacing $x$ with $x-x_o$ in the second formula. However, the convergence will only occur to the left or right of the singular point in considering the real interval of convergence. – James S. Cook Jun 22 '13 at 23:36

Your first solution is a power series. Your second solution is the Frobenius solution which allows for $r$ non-integer. There are 4 cases to consider for the second solution. The way in which the second solution is found differs slightly if:

1. you have distinct exponents which do not differ by an integer
2. you have distinct exponents which do differ by an integer
3. you have repeated exponents
4. you have complex exponents

The simplest example illustrating these is the Cauchy-Euler problem $ax^2y''+bxy'+cy=0$ which is the quintessential example of a second order ODE with a regular singular point.

To answer your question, if you seek a solution at an ordinary point use the power series solution. If you face a regular singular point then invoke the method of Frobenius.

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I know there is a different method for solving Euler-Cauchy equations, but if I didn't recognise a problem as being an Euler-Cauchy equation, could I use either the power series or frobenius method and still come up with the correct answer? – Sam Jun 23 '13 at 10:04
@Sam You look at $ar(r-1)+br+c=0$ for the Cauchy-Euler problem I mentioned in my post. This is the indicial equation from the Frobenius method. The wonderful thing about the Cauchy-Euler problem is that the rest of the Frobenius series vanishes. You just get the lowest order terms for the Frobenius solution. Sort of like solving $y^{n}(x)=0$ with a power series ansatz, it's a bit of an overkill since clearly an $n$-th order polynomial will do nicely. (well, just integrate $n$-fold times for that silly example) – James S. Cook Jun 23 '13 at 17:43