I have been stuck on this problem for days. Unfortunately, every three times I get it wrong, it regenerates a new matrix with new numbers and new initial conditions.
The new problem I have is:
Find the solution to the given system that satisfies the initial condition. $x'(t)=\begin{bmatrix}-3 & -1\\5 & -1\end{bmatrix}x(t)$, with a) $x(0)= \begin{bmatrix}-8 \\ 0\end{bmatrix}$, b) $x(\pi)= \begin{bmatrix}-1 \\ 1\end{bmatrix}$, c) $x(-2\pi)= \begin{bmatrix}5 \\ 1\end{bmatrix}$, d) $x(\frac{\pi}{4})= \begin{bmatrix}0 \\ 2\end{bmatrix}$
I think I'm getting the correct "base" equation, but I'm not 100% sure since it does not have us check it in their system.
Starting with the main matrix and naming it "A", I get $$A=\begin{bmatrix}-3 & -1\\5 & -1\end{bmatrix}$$ Therefore, $eig(A)=-2+2i$, $eig(A)=-2-2i$
Which results in the following (A-rI)(u)=0 $$\begin{bmatrix}-1+2i & -1\\5 & 1+2i\end{bmatrix}*\begin{bmatrix}u_1 \\ u_2\end{bmatrix}=\begin{bmatrix}0 \\ 0\end{bmatrix}$$ Doing row reduction, we get $$\begin{bmatrix}1 & 0.2+0.4i & u_1\\0 & 0 & u_2\end{bmatrix}$$ Therefore, $1u_1=-(0.2+0.4i)u_2$, or $u_1*\begin{bmatrix}1 \\ -0.2-0.4i\end{bmatrix}$
Resulting in the answer: $$x(t)=c_1*e^{-2t}[cos(2t)*\begin{bmatrix}1 \\ -0.2\end{bmatrix}-sin(2t)\begin{bmatrix}0 \\ -0.4\end{bmatrix}]+c_2*e^{-2t}[sin(2t)*\begin{bmatrix}1 \\ -0.2\end{bmatrix}+cos(2t)\begin{bmatrix}0 \\ -0.4\end{bmatrix}]$$
I plug in the initial conditions and get: $$x(0)=\begin{bmatrix}c_1 \\ -\frac{c1}{5}-\frac{2c_2}{5}\end{bmatrix}=\begin{bmatrix}-8 \\ 0\end{bmatrix}$$
But is this correct? I've done it this way before and kept on getting the problem wrong, and I've already spent almost six hours on it. I'm quite sure if someone pointed where I'm going wrong in this above process, I might be able to solve the other IVP's without any help.
As a side note, I'm checking my calculations that I do by hand in MATLAB.
Edit: Fixing the errors @Moo caught, I get the following equation: $$x(t)=e^{-2t}\begin{bmatrix}-c_1*cos(2t)-2c_1*sin(2t)-c_2*sin(2t)+2c_2*cos(2t) \\ 5c_1*cos(2t)+5c_2*sin(2t)\end{bmatrix}$$
Solving for $x(0)=\begin{bmatrix}-8\\0\end{bmatrix}$, I get $c_1=0, c_2=-4$, which results in $$x(t)=e^{-2t}\begin{bmatrix}4sin(2t)-8cos(2t)\\ -20sin(2t)\end{bmatrix}$$
Which is correct.
Following the same procedure I got b) correct as well. However, I've gotten c) wrong twice, and only have one more try left before I fail this part of the question.
For c, I get that $x(-2\pi)=\begin{bmatrix}5\\1\end{bmatrix}=e^{4\pi}\begin{bmatrix}-c_1+2c_2\\5c_1\end{bmatrix}$, which means that $c_1=e^{-4\pi}\frac{1}{5}$ and $c_2=e^{-4\pi}2\frac{3}{5}=e^{-4\pi}\frac{13}{5}$.
However, the answer I get by plugging these in for $c_1, c_2$: $$e^{-2t-4\pi}\begin{bmatrix}5cos(2t)-\frac{39}{5}sin(2t)\\cos(2t)+13sin(2t)\end{bmatrix}$$ is not correct.
Where might I be going wrong now?