I know how to find the eigenvalues and associated vectors, but I don't know how to finish off the problem when it involves complex values.
The eigenvalues are $\lambda = 2+i, 2-i$
Would someone kindly guide me through what to do next? I have been searching high and low online and offline to no avail.
So, the problem is as the title says, "Find the general solution to the inhomogenous system $y' = Ay + f$
where
A = $\begin{pmatrix} 3 & 2 \\ -1 & 1 \end{pmatrix}$ and $f= \begin{pmatrix} 0\\ e^{2t} \end{pmatrix}$
The answer is $y(t) = C_1e^{2t}\begin{pmatrix} sin (t) - cos (t) \\ cos (t) \end{pmatrix} + C_2e^{2t}\begin{pmatrix}-cos (t) - sin (t) \\ sin (t) \end{pmatrix}+e^{2t}\begin{pmatrix} 2 \\ -1 \end{pmatrix}$
Next step is to find the eigenvectors. In fact, all you need is an eigenvector for one of the eigenvalues. If, say, $u = \pmatrix{a + b i\cr c + d i\cr}$ (with $a,b,c,d$ real) is an eigenvector for $\lambda = 2 + i$, then its complex conjugate $\overline{u} = \pmatrix{a - b i\cr c - d i\cr}$ is an eigenvector for $\overline{\lambda} = 2 - i$.
And the real and imaginary parts of $u e^{\lambda t}$ are a fundamental set of (real) solutions of the homogeneous system. Notice that $$(a + b i)e^{(2 + i)t} = (a + b i) e^{2t} (\cos(t) + i \sin(t)) = ((a \cos(t) - b \sin(t)) + i (a \sin(t) + b \cos(t)))e^{2t}$$ so its real part is $(a \cos(t) - b \sin(t)) e^{2t}$ and its imaginary part is $(a \sin(t) + b \cos(t)) e^{2t}$.