# 2nd Order Differential Equation via Power Series

1. Oct 10, 2013

### VVS

1. The problem statement, all variables and given/known data
Problem 5 on the attached Sheet here View attachment M2Uebung2.pdf

2. Relevant equations
We studied the Power Series Method and how to calculate a linearly independent solution if one solution is already known.
So we need to find one solution (probably) using the power series method and then using that solution we can find the 2nd linearly independent solution.

3. The attempt at a solution

See the attachment for working. View attachment Übung 5.pdf
So I have found a recursion formula.
And it can be seen that all even coefficients above n=2 are vanishing, but what about a2 and the odd coefficients?

VVS

Last edited: Oct 10, 2013
2. Oct 10, 2013

### haruspex

It would appear that a0 and a1 can be arbitrarily assigned (unsurprising for the DE) and all further terms written in terms of them. So to get one solution you can plug in whatever you like for a0 and a1 (except both zero) and see what results. In fact, it's not hard to write out a general form for the odd terms using factorials.

3. Oct 10, 2013

### pasmith

In principle there are initial conditions on $y(0) = a_0$ and $y'(0) = a_1$.

Since you aren't given any initial conditions, you will have to choose some. Taking $a_0 = 1$ and $a_1 = 0$ seems good, since as you have observed all even terms other than $a_0$ and $a_2$ will vanish, as will all the odd terms. Then you can obtain the other solution by the suggested integral method.

4. Oct 11, 2013

### VVS

Hi Haruspex and Pasmith!

Thanks for your help. Yeah, I asked my professor whether there is something wrong with the assignement.
He just wrote that I should take a closer look.
I guess Pasmith is right. Any possible solution can be regarded as a solution for the differential equation.
So we can choose a0=1 and a1=0.
I will actually try to use the recursion formula and find a general expression for the second linearly independent solution.

VVS

5. Oct 11, 2013

### VVS

Second Linearly Independent Solution

Hey,

I got the solution for the first linearly independent solution of the Differential Equation.
But now I am facing another problem.

The integral for the second linearly independent turns out to be quite complicated.
I was able to break it down a bit by using integration by parts. See Working here View attachment Übung 5_3.pdf

But there must be simpler way.

thanks
VVS

6. Oct 11, 2013

### VVS

Actually we don't need to calculate the integral.
Sorry for wasting posts.
So this problem is solved.