Another series solution ODE problem

[TheFerruccio]

Homework Statement

Find a basis of solutions.

Homework Equations

(1-x^2)y''+(1-x)y'-3y = 0

The Attempt at a Solution

Using the series approach, having:

y=\sum_{n=0}^{\infty}a_nx^n

I ended up with an equation representing the coefficients for x^0

2a_2+a_1-3a_0 = 0

I'm confused because this is the first time I have encountered three coefficients for the same power of x. I just want to verify that I didn't make an algebra mistake, because I am used to finding that certain coefficients are linked to even powers, while others are linked to odd powers.

 

[LawlQuals]

Hi TheFerruccio,

Your work looks fine, I just went through the problem and obtain the same result you got. I have seen a lot of power series solutions that end up with three coefficients in their recursive relations, so it is no problem. This happens when you have second order differential equations. Given that a second order differential equation will admit two arbitrary constants for the general solution, your single equations in three unknowns is a way of saying there are two 'free parameters' (if I may borrow the term from linear algebra). Your final solution will be able to be expressed in terms of two solutions that depend each on one of the arbitrary constants: a_0 and a_1. If you had initial/boundary conditions, you would be able to find these constants.

For instance, the differential equation:

y'' + \omega^2y = 0

If you used a power series solution to find the result, you would get a recursive relation in three coefficients, and you could compute several terms. In the end you would be able to reduce all the coefficients to be put in terms the first two coefficients: a_0 and a_1, such that your final solution would look like:

y(t) = a_0\left(1 - \frac{(\omega t)^2}{2!} + \frac{(\omega t)^4}{4!} - \ldots\right) + a_1\left(\omega t - \frac{(\omega t)^3}{3!} + \frac{(\omega t)^5}{5!} - \ldots \right)

Which you would recognize as:

y(t) = a_0\cos\omega t + a_1\sin\omega t

That only happens though if your recursive relation would have three parameters in it to start with. It is a common situation though, so no worries.

 

[TheFerruccio]

Thanks for the help! I managed to put together some kind of solution, though it does not really match what was found using mathematica. I'm probably missing something with the algebra or the simplification. Here is the solution I have:

y = a_0(1 + \frac{3}{2}x^2 + \frac{1}{2}x^3 + \frac{3}{4}x^4 + \frac{3}{4}x^5 + \ldots) + a_1(x - \frac{1}{2}x^2 + \frac{1}{2}x^3 + \frac{1}{4}x^4 + \frac{1}{4}x^5 + \ldots)

EDIT: It looks like this works just fine. There is likely a simplification for some of the series, but I don't know it.