How to solve the homogeneous system of a differential equation I want to solve the system $\dot{x}=Ax$ where $ A = \left( \begin{matrix} 1 & 2 \\ 3 & 2 \end{matrix} \right).$ I found the eigenvalues of $A$ as $\lambda_1=4,\lambda_2=-1$ with correspongind eigenvectors $v_1=(2,3)^\intercal,v_2 =(1,-1)^\intercal $ respectively. Then, the solution of the system will be
$$x(t)=e^{At}x_0$$
so that I obtained $$x(t)=\left( \begin{matrix} \frac{3}{5}e^{-t} + \frac{2}{5}e^{4t} & -\frac{2}{5}e^{-t} + \frac{2}{5}e^{4t} \\ \frac{3}{5}e^{4t} + -\frac{3}{5}e^{-t} & \frac{3}{5}e^{4t} + \frac{2}{5}e^{-t} \end{matrix} \right) x_0$$
where $x_0$ is the initial condition. Now, if I say that $x_0 =(c_1,c_2) $ for some constants $c_1,c_2$ I obtain a sum which is very mixed as you can guess from the above matrix. (Note that I verified the matrix via Wolfram).
However, Paul's Online Notes and an online solver says that the solution can be found as
$$x(t)=c_1 e^{-t} (-1,1)^\intercal+ c_2 e^{4t}(2,3)^\intercal$$
that is not equal to the solution that I found. So, what is the problem with my solution?
Final note: The second solution takes the first eigenvector as $-v_1$.
 A: $$x(t)=\left( \begin{matrix} \frac{3}{5}e^{-t} + \frac{2}{5}e^{4t} & -\frac{2}{5}e^{-t} + \frac{2}{5}e^{4t} \\ \frac{3}{5}e^{4t} + -\frac{3}{5}e^{-t} & \frac{3}{5}e^{4t} + \frac{2}{5}e^{-t} \end{matrix} \right) x_0$$
$$x(t)=\left( \begin{matrix} \frac{1}{5}e^{-t}(3c_1-2c_2)  + \frac{2}{5}e^{4t} (c_1+c_2) \\ - \frac{1}{5}e^{-t}(3c_1-2c_2)  +  \frac{3}{5}e^{4t} (c_1+c_2) \end{matrix} \right) $$
$$x(t)=\left( \begin{matrix} e^{-t}k_1 +  {2}e^{4t} k_2 \\ - e^{-t}k_1  +  {3}e^{4t} k_2 \end{matrix} \right) $$
This agree with the answer you posted.
$$x(t)=k_1e^{-t}\left( \begin{matrix} 1 \\ - 1 \end{matrix} \right) +k_2e^{4t}\left( \begin{matrix}2 \\ 3\end{matrix} \right) $$
Where :$$k_1= \frac {3c_1-2c_2 }5, k_2= \frac {c_1+c_2}5 $$
There is nothing wrong with your solution.
And for initial conditions $(x_0)$ you have:
$$x(0)=k_1\left( \begin{matrix} 1 \\ - 1 \end{matrix} \right) +k_2\left( \begin{matrix}2 \\ 3\end{matrix} \right)$$
You have the same formula as the one in Paul's notes.

found the eigenvalues of $A$ as $λ_1=4,λ_2=−1$ with correspongind eigenvectors $v_1=(2,3)^t,v_2=(1,−1)^t$ respectively. Then, the solution of the system will be:
$$x(t)=c_1e^{4t}v_1+c_2e^{-t}v_2$$
