# Complex-valued solutions to real-valued solutions

1. Dec 3, 2005

### NINHARDCOREFAN

I'm going crazy trying to figure out how to do this. Here are two eigen value solutions: r=-1+-2i to a problem. And the two eigen vectors are (-2i, 1) and (2i, 1). How do I covert this to a real-valued solution? The book used (-2i, 1) to come up with the solution x = c * e^(-t)( -2sin(2t), cos(2t) ) + d * e^(-t)( 2cos(2t), sin(2t) )
c and d are constants.
I know that real value of r=1+-2i should be e^(-t)sin(2t) + e^(-t)cos(2t) but I don't how they came up with their answer using this fact.

2. Dec 3, 2005

### HallsofIvy

Staff Emeritus
$$e^{ix}= cos(x)+i sin(x)$$. Does that help?

Last edited: Dec 6, 2005
3. Dec 5, 2005

### jim mcnamara

$$e^{ix}= cos(x)+i sin(x)$$