# Express the general solution?

1. Sep 15, 2013

### Success

Express the general solution of x'=(1, 2, 3, 0, 1, 2, 0, -2, 1)x in terms of real-valued functions. (this is 3x3 matrix, 1, 2, 3 on the left, 0, 1, 2 in the middle, 0, -2 and 1 on the right. I found that the roots are 1, 1+2i, 1-2i. And a=2, b=-3, c=2 for the first root. a=0, b=1, c=i for the second root. a=0, b=-1, c=i for the third root. but how do I get the answer?)

2. Sep 15, 2013

### PhysicsandSuch

Your general solution is just the sum of each independent solution by superposition with constants to be found from initial conditions. Build each independent solution from the eigenvector multiplied by the exponential of the eigenvalue multiplied with the dependent variable.

For example, assuming $x = x(t)$, $X = \sum_i A_i ψ_i e^{λ_i t}$ for constants $A_i$ and eigenvectors $ψ_i$ associated with eigenvalues $λ_i$.

Then just rearrange your solutions into real valued functions like the questions asks. (It looks like the solutions take the form of exponentials, or exponentials multiplied with sins and cos at first glance).

3. Sep 16, 2013

### Success

But how do I get the answer? The answer is x=c1*(2, -3, 2)e^t+c2*e^t*(0, cos(2t), sin(2t))+c3*e^t(0, sin(2t), -cos(2t)).

4. Sep 16, 2013

### PhysicsandSuch

Right, so I pretty much gave you an explicit roadmap to get that exact answer.

You are going to have to do some more work and at least try to work it out using what I've already told you. (You are 75% there already).

If you have more questions, post how far you get and I'll be happy to answer your questions about where you are getting stuck, but continuing to ask "how do I get the answer" isn't going to get you very far, I'm afraid.