Differential equation system with complex eigenvalues I need to solve this equation 
$$ x'=x+y,     y'=-2x+3y$$
I get the matrix rigor and the eigenvectors complex $2-i$ and $2+i$.  When I try to apply the eigenvectors associated the solution for $x$ I simply can't find ( I do find the $y$ though).
The solution on the book for $x=e^{2t}(a\cos(t)+b\sin(t))$
Any help would be appreciated.
 A: For the system:
$$ x'=x+y,~y'=-2x+3y$$
We have:  $A = \begin{bmatrix} 1 & 1\\ -2 & 3 \end{bmatrix}$.
As you found, setting up $|A- \lambda I| = 0$ and solving for the eigenvalue/eigenvector pairs yields:


*

*$\lambda_1 = 2+i, ~v_1 = (\frac{1}{2}(1-i), 1)$   

*$\lambda_2 = 2-i, ~v_2 = (\frac{1}{2}(1+i), 1)$   


So, for the first eigenvalue, we have:
$e^{\lambda_1 t}v_1 = e^{(2+i)t}\begin{bmatrix}\frac{1}{2}(1-i)\\1\end{bmatrix} = e^{2t}e^{it}\begin{bmatrix}\frac{1}{2}(1-i)\\1\end{bmatrix} = e^{2t}(\cos t + i \sin t)\begin{bmatrix}\frac{1}{2}(1-i)\\1\end{bmatrix} = \begin{bmatrix} \frac{1}{2}e^{2t}(\cos t + \sin t) + \frac{1}{2} i (\sin  t-\cos t))\\ e^{2t}(\cos t + i \sin t) \end{bmatrix} $
So, our solution can be written as (because we know that the real and imaginary parts are both independent solutions):
$$W(t) = \begin{bmatrix}x(t)\\ y(t) \end{bmatrix} = a e^{2t}\begin{bmatrix}\frac{1}{2}(\sin t + \cos t)\\ \cos t \end{bmatrix} + b e^{2t}\begin{bmatrix}\frac{1}{2} (\sin t - \cos t)\\ \sin t \end{bmatrix}$$
If you want to write these out independently, we have:


*

*$\displaystyle x(t) = \frac{1}{2}e^{2t}\left( a(\sin t + \cos t) + b(\sin t - \cos t)\right)$

*$\displaystyle y(t) = e^{2 t}(a \cos t + b \sin t)$


Did you swap the your labels for $x(t)$ and $y(t)$?
A: Let's rewrite your equations as follows,assuming that $Dx=x',~Dy=y'$:
$$ \left\{
        \begin{array}{ll}
            (D-1)x-y=0 \\
            2x+(D-3)y=0
        \end{array}
    \right.
$$
By solving this homogenous system of equations, we have:
$$ \left\{
        \begin{array}{ll}
            (D-1)(D-3)x+2x=0 \\
            (D-1)(D-3)y+2y=0
        \end{array}
    \right.
$$ which is equivalent to the following system:
$$ \left\{
        \begin{array}{ll}
            x''(t)-4x'(t)+5x(t)=0 \\
            y''(t)-4y'(t)+5y(t)=0
        \end{array}
    \right.
$$ Now, I think you can do the rest.
