# Creating series solutions for a non-constant coefficient ODE

1. May 23, 2016

### rmiller70015

1. The problem statement, all variables and given/known data
This is for differential equations with nonconstant coefficients and I wasn't so great at series and sequences in calculus so when I came across this example problem I wasn't sure how they got to their final form. If someone could explain it to me that would be really helpful.

2. Relevant equations
The ODE is:
$$y"-2xy+2y=0$$

3. The attempt at a solution
We assume there is some solution of the form:
$$y(x)=\sum_{n=0}^{\infty}a_nx^n$$
The derivatives are:
$$y'(x)=\sum_{n=1}^{\infty}na_{n}x^{n-1}\:\: \text{and}\:\: y''(x)=\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}$$
A change of index gives
$$y'(x)=\sum_{n=0}^{\infty}na_nx^{n-1}\:\:\text{and}\:\:y''(x)=\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^n$$
Putting these into the ODE gives:
$$\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n-2x\sum_{n=0}^{\infty}na_nx^{n-1}+2\sum_{n=0}^{\infty}a_nx^n=\sum_{n=0}^{\infty}\Big((n+2)(n+1)a_{n+2}-2na_n+2a_n\Big)x^n=0$$
The Identity Theorem gives:
$$(n+2)(n+1)a_{n+2}-2na_n+2a_n=0$$
Solving this for $a_{n+2}$ gives:
$$a_{n+2}=\frac{2(n-1)a_n}{(n+2)(n+1)}$$
Odd coefficients are zeros except for $a_1$ because:
$$a_3=\frac{2(1-1)a_1}{(n+2)(n+1)}=0\:\: \text{and all other coefficients will depend on a multiple of}\:a_3$$
The even coefficients are given by:
$$a_{2k+2}=\frac{2(2k-1)a_{2k}}{(2k+2)(2k+1)}$$
This is a worked example from the book and the answer is:
$$\frac{2(2k!)}{k!}$$

I'm not sure how they got this final form, but I think it has something to do with the even number recursion formula, which can be written as:
$$a_{2k+2}=\frac{2(2k-1)a_{2k}}{(2k+2)(2k+1)}=\frac{2(2k-1)}{(2k+2)(2k+1)}\frac{2(2k-3)a_{2k-2}}{2k(2k-1)}$$
This is where I am having the problem. The book says the $a_{2k+2}$ can be written as:
$$\frac{2^{k+1}(2k-1)(2k-3)...3\cdot2\cdot1}{(2k+2)!}a_0$$
I understand where the $2^{k+1}$ and $a_0$ are coming from, but the other terms I am unsure of.

2. May 24, 2016

### Buzz Bloom

Hi rmiller:

I have not yet digested all of your calculations, but I noticed a typo in your ODE. You are missing an apostrophe. The second term should be
-2xy' .​

Since the equation is linear, any constant multiple of a solution is also a solution. This means that a0 is an arbitrary constant.
Since this is a second order equation, there should be two independent solutions.

It has been decades since I worked with ODEs so I may be mis-remembering the following.

I think you can also choose a1 to be another arbitrary constant. Then find how a2 relates to a1 and a0. You then need a recursion relationship between
a2k+2 and a2k.​

The solution given
a2k = 2(2k!)/k!​
seems to be based on assuming that a0 = 4 and ignoring a1.

You should be able to find the second solution by assuming a0 = 0 and a1 = 1.

I notice that the solution given above is inconsistent with the previous even coefficients expression for a2k+2.
Assuming a0 = 1, the expression for a2k+2 calculates
a2 = -a0.​
However, the book answer 2(2k!)/k! is always positive.

Hope that this helps.

Regards,
Buzz

Last edited: May 24, 2016