# Complex root for characteristic equation

1. Dec 19, 2012

### aaaa202

Suppose your characteristic equation for the 2nd order equation has complex roots
r+ and r-

These are conjuagtes of each other so the general solution is:

y = Aer+ + Ber-

My book chooses the constants A and B as conjugates of each other for the reason that this constructs a real solution (not very hard to see if you plug A = E + iF and B = E-iF into the equation above).

But my question is: How does my book know that this constructs all possible solutions that are real?

2. Dec 19, 2012

### Dick

You forgot to put the independent variable in the exponents. But it's because the general solution has two constants that determine the solution. If y(t) is the solution it's determined by y(0) and y'(0). Your solution also has two constants, E and F. If you know E and F you can find y(0) and y'(0) and vice versa.

Last edited: Dec 19, 2012
3. Dec 20, 2012

### HallsofIvy

The basic theory of "linear homogeneous differential equations of order n" (with real coefficients) is that the set of all solutions forms a vector space, over the real numbers, of dimension n. As long as you have n independent solutions, the general solution can be written as a linear combination of them.