Question

Could someone please help me with solving this second order series example:
y''-xy'+2y=0.

It must be solved by power series, with a center a, finding the recurrent formula, and two linearly independent solutions. Does anyone know of a good Diff EQ book-the one I have skips slot of information in the explanation, and that makes it difficult to follow.

Answer 1

I'm gonna try to attempt this mainly because i was working on this same problem the other day, and I agree with you there aren't too many textbooks that do a good job of covering higher order differential equations with variable coefficients, but I'll give it a shot.

First, things first. We're going to use a power series substitution to solve this equation. We need to let\[y=\sum_{n=0}^{\infty}a_{n}x^{n}\]
and we'll need two derivatives:
\[y'=\sum_{n=1}^{\infty}a_{n}nx^{n-1}\]
\[y''=\sum_{n=2}^{\infty}a_{n}n(n-1)x^{n-2}\]
Then we're going to substitute these into the differential equation:
\[y''-xy'+2y=\sum_{n=2}^{\infty}a_{n}n(n-1)x^{n-2}-x\sum_{n=1}^{\infty}a_{n}nx^{n-1}+2\sum_{n=0}^{\infty}a_{n}x^{n}\]

which simplifies to:

\[\sum_{n=2}^{\infty}a_{n}n(n-1)x^{n-2}-\sum_{n=1}^{\infty}a_{n}nx^{n}+\sum_{n=0}^{\infty}2a_{n}x^{n}\]

Now this is where it gets hard.
The first step is to find the exponent of x of the FIRST term of each series. Example the first series starts at n = 2 and the exponent in x is n-2 so the first term has power 0. Repeating the process, the second series has first exponent 1 and the third series has first exponent 0. These all need to be have the same first term exponent. This is adjusted by "pulling the terms" out of the series (or computing a term and adjusting the series) pull the terms out of the smaller exponent series. The first term of the first series (by plugging in n=2) is
\[2c _{2}x ^{0}\]
and the first term of the last series is
\[c _{0}x^{0}\]
"Pulling out" these terms adjusts the powerseries equation to look like this:
\[2c _{2}x ^{0}+\sum_{n=3}^{\infty}a_{n}n(n-1)x^{n-2}-\sum_{n=1}^{\infty}a_{n}nx^{n}+c _{0}x^{0}+\sum_{n=0}^{\infty}2a_{n}x^{n}\]
Now we can adjust the first series by "shifting the index n" to n=1. Then we can combine everything into one series.
\[\sum_{n=3}^{\infty}a_{n}n(n-1)x^{n-2}=\sum_{n=1}^{\infty}a_{n+2}(n+2)(n+1)x^{n}\]
(so if n = 3 becomes n=1, every n in the series section is increased by 2)
Now we can combine everything to one big happy series
\[c_{0}+2c _{2}+\sum_{n=1}^{\infty}[a_{n+2}(n+2)(n+1)-a_{n}n+2a_{n}]x^{n}=0\]
The recursive formula is now solved where the series equals 0:
\[a_{n+2}(n+2)(n+1)-a_{n}n+2a_{n}=0\]
\[a_{n+2}(n+2)(n+1)=a_{n}n-2a_{n}\]
\[a_{n+2}=\frac{ (n-2) }{ (n+2)(n+1)}a_{n}\]

This is the formula to find the values of the series of the solution. It will only have odd terms because c_{0} and c_{2} both equal 0. Hope this helps. (And sorry for any mistakes in advance, typing this wasn't a whole bunch of fun!)

Answer 2

whoops made a mistake with the coefficents in the series. it should be
\[2c_{0}x^{0} \]
when you pull out the first term of the third series. So after combining the terms of the series it should look like:
\[2c_{0}x^{0}+2c_{2}x^{0}+\sum_{n=1}^{\infty}[a_{n+2}(n+1)(n+2)-a_{n}n-2a_{n}]=0\]

This means that

\[2c_{0}+2c_{2}=0\]
and
\[c_{2}=-c_{0}\]
thus
\[y=(1-x^2)c_0\]
is one solution of the equation and the odd series starting at n=1 is the other. My apologies.