Series solution of a differential equation

[spaghetti3451]

Homework Statement

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

Homework Equations

The Attempt at a Solution

[itex]
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)!}
[/itex][itex]
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)!} \\
[/itex]

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

Could you please help me out?

 

[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.

 

[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:

[itex]
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)!} \\
[/itex]

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

[itex]
C\ = a_{0}\ and\ D\ = \frac{a_{1}}{2}.
[/itex]

What do you think?

 

[Fightfish]

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