General Solution of ODE (complex eigenversion) I am trying to figure out the general solution to the following matrix:
$ \frac{d\mathbf{Y}}{dt} = \begin{pmatrix} -3 & -5 \\ 3 & 1 \end{pmatrix}\mathbf{Y}$
I got a solution, but it is so complex I am not sure, if it's even remotely right....:
My Solution:
$x = 1/3*k_1*(-\sqrt{11}*cos(\sqrt{11}*t))-2*sin(\sqrt{11}*t)+1/3*k_2*(\sqrt{11}*sin(\sqrt{11}*t)-2*cos(\sqrt{11}*t))$
$y = k_1*sin(\sqrt{11}*t)+k_2*cos(\sqrt{11}*t)$
After that my task is to solve the inital-value problem for the same matrix with:
$Y_0=(4, 0)$
I have an idea where to start, should I find the correct solution to the general solution problem, but it looks so complex, that it overwhelms me...
 A: The coefficient matrix $A$ has characteristic polynomial
\begin{align}
        \left|\begin{array}{cc}
              \lambda+3 & 5 \\
                -3 & \lambda-1
              \end{array}\right|
      & = (\lambda+3)(\lambda-1)+15 \\
      & = \lambda^{2}+2\lambda+12 \\
      & = (\lambda+1)^{2}+11 \\
      & = (\lambda+1+i\sqrt{11})(\lambda+1-i\sqrt{11}).
\end{align}
So the eigenvalues of the coefficient matrix are $-1\pm i\sqrt{11}$. There is one eigenvector correspondending to each eigenvalue. If $A$ is the coefficient matrix in the differential equation, then
$$
             A-(-1+i\sqrt{11})I=\left[\begin{array}{cc}
                                       -2-i\sqrt{11} & -5 \\
                                       3 & 2-i\sqrt{11}
                                      \end{array}\right]
$$
gives an eigenvector
$$
                  X_{+} = \left[\begin{array}{c}
                                 -5 \\ 2+i\sqrt{11}
                                \end{array}\right]
$$
Similarly,
$$
               A-(-1-i\sqrt{11})I = \left[\begin{array}{cc}
                                          -2+i\sqrt{11} & -5 \\
                                            3 & 2+i\sqrt{11}
                                          \end{array}\right]
$$
gives an eigenvector
$$
               X_{-}= \left[\begin{array}{c}
                       -5 \\
                       2-i\sqrt{11}
                      \end{array}\right]
$$
So the general solution of the equation is
$$
          P(t) = \alpha e^{-t+i\sqrt{11}t}X_{+} + \beta e^{-t-i\sqrt{11}t}X_{-}
$$
The solution for $Y_{0}=(4,0)$ is obtained by solving for $\alpha$, $\beta$ such that
$$
                \alpha\left[\begin{array}{c} -5 \\ 2+i\sqrt{11}\end{array}\right]+\beta\left[\begin{array}{c} -5 \\ 2-i\sqrt{11}\end{array}\right]=\left[\begin{array}{c} 4 \\ 0 \end{array}\right] \\
          \left[\begin{array}{c} \alpha \\ \beta \end{array}\right]
   = \frac{1}{10i\sqrt{11}}\left[\begin{array}{cc}
                         2-i\sqrt{11} & 5 \\ 
                        -2-i\sqrt{11} & -5\end{array}\right]\left[\begin{array}{c}4 \\ 0\end{array}\right]
     = \frac{2}{5i\sqrt{11}}\left[\begin{array}{c}2-i\sqrt{11} \\ -2-i\sqrt{11}\end{array}\right]
$$
Therefore,
$$
    P(t) = \frac{2(2-i\sqrt{11})}{5i\sqrt{11}}e^{-t+i\sqrt{11}t}X_{+}
           -\frac{2(2+i\sqrt{11})}{5i\sqrt{11}}e^{-t-i\sqrt{11}t}X_{-}.
$$
The first coordinate of $P(t)$ is
$$
        -e^{-t}\left[\frac{4}{i\sqrt{11}}\{e^{i\sqrt{11}t}-e^{-i\sqrt{11}t}\}\right]+e^{-t}\left[2\{e^{i\sqrt{11}t}+e^{-i\sqrt{11}t}\}\right] \\
    = -\frac{8}{\sqrt{11}}e^{-t}\sin(\sqrt{11}t)+4e^{-t}\cos(\sqrt{11}t).
$$
The second coordinate of $P(t)$ is
$$
   \frac{2(2-i\sqrt{11})(2+i\sqrt{11})}{5i\sqrt{11}}e^{-t+i\sqrt{11}t}
   -\frac{2(2+i\sqrt{11})(2-i\sqrt{11})}{5i\sqrt{11}}e^{-t-i\sqrt{11}t} \\
  = \frac{12}{\sqrt{11}}e^{-t}\sin(\sqrt{11}t)
$$
