# Homework Help: Differential Equations

1. Dec 8, 2009

### Unto

The general solution of the second order DE

$$y'' + a^2y = 0$$

is

$$y = A cos ax + B sin ax$$

WTF?

Someone explain why?

2. Dec 8, 2009

### Staff: Mentor

The characteristic equation is r2 + a2 = 0, which has solutions r = +/- ai. This means that the general solution is all linear combinations of {eiax, e-iax} = {cos ax + i sin ax, cos ax - i sin ax). By taking suitable linear combinations of these two complex functions, you can get two real solutions, sin(ax) and cos(ax).

3. Dec 8, 2009

### Unto

Why does the linear combinations include complex conjugate?

4. Dec 8, 2009

### Staff: Mentor

A 1st order differential equation has 1 basic solution, a 2nd order differential equation has 2 basic solutions, a 3rd order differential equation has 3 basic solutions, and so on.

The basic solutions are all of the form erx, where r is a root of the characteristic equation. If the characteristic equation has complex roots, these roots always come in pairs - the complex conjugates. For the sake of convenience, instead of writing e(a + bi)x and e(a - bi)x, we do a little algebra and write these functions as eaxcos(bx) and eaxsin(bx) so that we don't have to mess with imaginary numbers at all.

5. Dec 8, 2009

### Unto

Yes I understand that, I am proficient with complex numbers.

6. Dec 8, 2009