I have the following linear differential equation system:
$$x' = A x$$ where $$ A = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 3 & 1 & -2 \\ 2 & 2 & 1 \end{array} \right) $$
I have computed the eigenvalues and eigen vectors, which are:
$$\lambda_1 = 1 \qquad \qquad v_1 =(2,-2,3)$$ $$\lambda_2 = 1 + 2i \qquad \qquad v_2 = (0,-i,1)$$ $$\lambda_3 = 1 - 2i \qquad \qquad v_3 = (0,i,1)$$
Now, the theory on my book gives the following functions as a base of the solutions of the linear system:
$$ \phi_i(t) = e^{t \lambda_i} \sum_{i=0}^{n_i-1} \dfrac{t^k}{k!}(A - \lambda_i I_n)^k v_i$$
where $n_i$ is the multiplicity of each eigenvalue. Using this formula, I get the following functions as a base of solutions from my problem:
$$\phi_1(t) = e^t (2,-2,3) \qquad \phi_2(t) = e^{(1+2i)t} (0,-i,1) \qquad \phi_3(t) = e^{(1-2i)t}(0,i,1)$$
Nevertheless, I am interested only in real solutions. I have noticed than $\phi_2(t),\phi_3(t)$ are conjugates, so i think adding them or sustracting them (and iI think I should be able to do so without problems, given that those are a base of a vector space) which would give me the real solution I am interesting in in form of sines and cosines.
Am I right? How would I do that in this particular example?
You can indeed take $(\phi_1,\phi_2+\phi_3,i^{-1}(\phi_2-\phi_3))$ for a base of real solutions.