Power series for singular DEs

Mathematics 1512 Honors Calculus II Fall, 2000

In last week's quiz, you were asked to solve the differential equation

A couple of you tried to solve this with power series, but ran into difficulties. (We are sorry that we did not warn you off this method, which is usually a good one, but not the best one for this problem.)

What was the problem, and what can be done about it? The cause of the problem is the factor x² in front of the derivative, which made it tricky to solve for the power-series coefficients when you plug in. If you did solve the equation with an integrating factor, you found a function which blows up at x=0, so in fact it does not have a Taylor series.

You can still solve the problem with power series, if you are careful. Here is how it goes:

Suppose that y=sum of a_n x^n, so that $x^2 y=sum of a_n n x^{n+1}.$ Notice I don't say what power the series begins with. Next plug:

$sum of a_n n x^{n+1} + sum of a_n 3 x^{n+1} = 1 = x^0.$

Now, the only way we can have x⁰ on the left is to allow n=-1, so let's do that. The coefficient a_-1 solves

_-1

so a_-1 = 1/2. Next we ask about the other a_n. If we equate powers we find that a_n solves

_-1

so a_n = 0, except when n=-3, which is not determined. Our solution is therefore

$y=(1/2) x^{-1} + a_3 x^{-3}.$

This is the correct general solution.