# Thread: Power series solutions to ODE

1. ## Power series solutions to ODE

In an attempt to find a power series solution to an ordinary differential equation, our professor illustrated the concept with an example. As a try-at-home exercise, the following homogeneous equation was given:

y''-5y'+6y=0

The above differs from his example critically - a_k+2 is in terms of a_k+1 as well as a_k which yields a2,a3,a4... in terms of a1 & a0.

My calculatios are here:

http://johnny.sas.upenn.edu/~gsobti/power.htm

The next step of separing a0 & a1 terms is proving to be erroneous.

Would aprpeciate some assistance.

Best,
wirefree

2. It's a second order linear ODE so it's going to have two linearly independent solutions, and the general solution will be a linear combination of these. Probably the easiest way to handle the recurrence relation is to get the two lin. ind. solutions as follows: for the first one, take a_0=1 and a_1=0. For the second one, take a_0=0 and a_1=1. That will have the effect of "separating the a_0 and a_1 terms", as you wanted. You will get two solutions, and you can then form a lin. comb. of these to get the general solution.

3. Originally Posted by Opalg
Probably the easiest way to handle the recurrence relation is to get the two lin. ind. solutions as follows: for the first one, take a_0=1 and a_1=0. For the second one, take a_0=0 and a_1=1. That will have the effect of "separating the a_0 and a_1 terms", as you wanted.
Appreciate the response, Opalg.

As suggested, I proceeded with successively equating a_0 & a_1 to zero and calculating the resulting series.

The resulting series, however, does not resemble the exponential series for 3x & 2x as the two solutions are found to be using characteristic equation method. Here are the results:

http://johnny.sas.upenn.edu/~gsobti/1.htm

The polynomial used is:

http://johnny.sas.upenn.edu/~gsobti/2.htm

4. I see what you mean. (I hadn't even noticed that the diff. equation has an easy analytical solution. In my defence, I never claimed that the recurrence relation would lead to an easy solution.)

If you want to retrieve the exponential solution from the recurrence relation $\displaystyle (n+2)(n+1)a_{n+2} -5(n+1)a_{n+1} +6a_n=0$, I would start by noticing that if you write $\displaystyle b_n = n!a_n$ then the relation becomes $\displaystyle b_{n+2}-5b_{n+1}+6b_n=0$. You solve this difference equation by the standard auxiliary equation method to get $\displaystyle b_n=A.2^n+B.3^n$. (Of course, that method is far more laborious than solving the diff. eqn directly by using the same auxiliary equation.)