# Series solution of a differential equation

1. Jun 5, 2013

### spaghetti3451

1. The problem statement, all variables and given/known data

Solve the following differential equation by power series and also by an elementary method. Verify that the series solution is the power series expansion of your other solution.

y'' = - 4y

2. Relevant equations

3. The attempt at a solution

$y'' = - 4y \\ \frac{d^{2}y}{dx^{2}} + 4y = 0 \\ Characteristic\ equation : λ^{2} + 4 = 0 \Rightarrow λ = ± 2i \\ y = A\cos(2x) + B\sin(2x) \\ y = A\ \sum^{m = ∞}_{m = 0} (-1)^{m} \frac{(2x)^{2m}}{(2m)!} + B\ \sum^{m = ∞}_{m = 0} (-1)^{m} \frac{(2x)^{2m+1}}{(2m+1)!}$

$Let\ y\ = \sum^{n = ∞}_{n = 0} a_{n}x^{n} \\ \sum^{n = ∞}_{n = 0} n(n-1)a_{n}x^{n-2} + 4\sum^{n = ∞}_{n = 0} a_{n}x^{n} = 0 \\ \sum^{n = ∞}_{n = 2} n(n-1)a_{n}x^{n-2} + 4\sum^{n = ∞}_{n = 0} a_{n}x^{n} = 0 \\ \sum^{n = ∞}_{n = 2} n(n-1)a_{n}x^{n-2} + 4\sum^{n = ∞}_{n = 2} a_{n-2}x^{n-2} = 0 \\ \sum^{n = ∞}_{n = 2} [ n(n-1)a_{n} + 4 a_{n-2} ] x^{n-2} = 0 \\ n(n-1)a_{n} + 4 a_{n-2} = 0,\ n = 2, 3, 4, ... \\ a_{n} = - \frac{(2)^{2}}{n(n-1)} a_{n-2},\ n = 2, 3, 4, ... \\ a_{2m} = - \frac{(2)^{2}}{(2m)(2m-1)} a_{2m-2}\ and\ a_{2m+1} = - \frac{(2)^{2}}{(2m+1)(2m)} a_{2m-1},\ m = 1, 2, 3, ... \\ a_{2m} = (-1)^{m} \frac{(2)^{2m}}{(2m)!} a_{0}\ and\ a_{2m+1} = (-1)^{m} \frac{(2)^{2m}}{(2m+1)!} a_{1},\ m = 1, 2, 3, ... \\ y = a_{0} \sum^{m = ∞}_{m = 0} (-1)^{m} \frac{(2x)^{2m}}{(2m)!} + a_{1} \sum^{m = ∞}_{m = 0} (-1)^{m} \frac{(2)^{2m}(x)^{2m+1}}{(2m+1)!} \\$

So, the solutions don't match because of the missing factor of 2 in the second summation of the series solution.

Last edited: Jun 5, 2013
2. Jun 5, 2013

### Fightfish

a0, a1, A and B are as-yet undetermined constants that come from the initial or boundary conditions. They can and will happily absorb any constant factors.

3. Jun 5, 2013

### spaghetti3451

Yeah, I figured this out just before I logged on to read your post.

Anyway, here's the corrected solution:

The last line of my solution could be written as follows:

$y = a_{0} \sum^{m = ∞}_{m = 0} (-1)^{m} \frac{(2x)^{2m}}{(2m)!} + \frac{a_{1}}{2} \sum^{m = ∞}_{m = 0} (-1)^{m} \frac{(2x)^{2m+1}}{(2m+1)!} \\$

so that a comparison with the solution obtained via the elementary method will lead us to conclude that

$C\ = a_{0}\ and\ D\ = \frac{a_{1}}{2}.$

What do you think?

4. Jun 5, 2013

### Fightfish

You mean A and B instead of C and D I guess. Yup, otherwise looks fine.