Question

Second Order Ordinary Differential Equation: Frobenius Method

Answer 1

\[xy'' + 2 y'+xy = 0 \]
\[x(y''+y)+2y = 0\]
and so x = 0 is regular oridinary point.

Answer 2

\[y = \sum a_m x^m \]
\[y' = \sum ma_m x^{m-1}\]
\[y'' = \sum m(m-1)a_mx^{m-2} \]
substitute into the ODE and then I got:
\[\sum m(m-1)a_m x^{m-1} +2 \sum ma_m x^{m-1} + x \sum a_m x^m = 0\]

Answer 3

let k = m-1, m = k+1, and so I have:
\[\sum k(k+1)a_mx^k = \sum m(m+1)a_mx^m - \sum m(m-1)a_mx^m \]
I shift k to m, but then I am not sure about the last term is necessary or simply it's just misintepration.

Answer 4

@phi can you help me with this. I just got stuck with shifting the k to m and having problems to determine the indicial equation.

Answer 5

I think part of the problem is you're not keeping track of your indices which because of this extra factor out front, it changes things quite a bit.

So specifically they want you to use this as your solution:
\[\large y = x^r \sum_{m=0}^\infty a_m x^m\]

So maybe to make your life easier, pull it into the sum as:

\[\large y = \sum_{m=0}^\infty a_m x^{m+r}\]

In the next step they are wanting you to pick r=-1 so probably worth noting your series will look like this eventually, which is subtly different from the ordinary power series you're used to.

\[y=\frac{a_0}{x}+a_1+a_2x+a_3x^2+\cdots\]

Your method seems like the right way to go though.

Answer 6

\[y = \sum a_m x^{m+r}\]
\[y' = \sum (m+r)a_mx^{m+r-1} \]
\[y'' = \sum (m+r)(m+r-1)a_mx^{m+r-1} \]
Substitute into ODE,
\[\sum (m+r)(m+r-1)a_mx^{m+r-1} + 2 \sum (m+r)a_m x^{m+r-1} + \sum a_m x^{m+r+1}\]
and now what do I do with them... I am lost for this part as it seems that I will have to find the indicial equation to solve this