Solving ODE-systems with Matrix exponential is wrong? Originally I've learned that the solution of a systems of coupled ODE:
$$\underbrace{\left[\begin{array}{cc}{y_1}'(x)\\ \vdots \\{y_n}'(x)\end{array}\right]}_{y'(x)}= 
\underbrace{\left[\begin{array}{cccc}&a_{1\,1} &\cdots &a_{1\,n}
\\ &\vdots \quad &&\vdots \\
&a_{n\,1}&\cdots&a_{n\,n}\end{array}\right]}_{A}\,
\underbrace{\left[\begin{array}{cc}{y_1}(x)\\ \vdots \\{y_n}(x)\end{array}\right]}_{y(x)}$$
is determined by: $$y(x) = \exp(A\,x)\,C$$ where $C$ is a vector with constants $\left[\begin{array}{cc}C_1\\ \vdots \\C_n\end{array}\right]$ and $\exp(A\,x)$ the matrix exponential, that can be at best calculated by: $$\exp(A\,x) = V^{-1}\,\exp(\Lambda\,x)\,V$$ where $V$ is a vector full of Eigenvectors: $\left[\begin{array}{cc}v_1&\cdots&v_n\end{array}\right]$
and $\Lambda$ a matrix full of Eigenvalues on its main diagonal: $\left[\begin{array}{ccc}\lambda_1&\\ &\ddots\\&&v_n\end{array}\right]$
Now apparently this leads to another solution compared to: $$y(x) = c_1\,v_1\,\exp(\lambda_1\,x)+\cdots+c_n\,v_n\,\exp(\lambda_n\,x)$$
Even if one told me both solutions were to solve a system of ODE

For example consider the system:
$$\left(\begin{array}{cc}{y_1}'(x) \\ {y_2}'(x)\end{array}\right) = \left(\begin{array}{cc} 4 & 10\\ 8 & 2 \end{array}\right)\,\left(\begin{array}{cc}{y_1}(x) \\ {y_2}(x)\end{array}\right)$$
with Eigenvalues $\lambda_1 = 12, \lambda_2 = -6$ and Eigenvectors: $v_1 = \left(\begin{array}{cc}1 \\ 8/10\end{array}\right), v_2 = \left(\begin{array}{cc}1 \\ -1\end{array}\right)$
According to the second plain solution process I'd obtain:
$$\left(\begin{array}{cc}{y_1}(x) \\ {y_2}(x)\end{array}\right) = \left(\begin{array}{cc}1 \\ 8/10\end{array}\right)\,\exp(12\,x)+\left(\begin{array}{cc}1 \\ -1\end{array}\right)\,\exp(-6\,x)$$
However the matrix exponential spits:
$$\left(\begin{array}{cc}{y_1}(x) \\ {y_2}(x)\end{array}\right) = C\,\left(\begin{array}{cc} \frac{4\,{\mathrm{e}}^{-6\,x}}{9}+\frac{5\,{\mathrm{e}}^{12\,x}}{9} & \frac{5\,{\mathrm{e}}^{12\,x}}{9}-\frac{5\,{\mathrm{e}}^{-6\,x}}{9}\\ \frac{4\,{\mathrm{e}}^{12\,x}}{9}-\frac{4\,{\mathrm{e}}^{-6\,x}}{9} & \frac{5\,{\mathrm{e}}^{-6\,x}}{9}+\frac{4\,{\mathrm{e}}^{12\,x}}{9} \end{array}\right)\,$$
Probably those two are inconvenient, because the constants are set differently. In fact the second approach is independent of constants somehow. So how's that all in relation with each other?

We have
$$\left(\begin{array}{cc}{y_1}'(x) \\ {y_2}'(x)\end{array}\right) = \left(\begin{array}{cc} 4 & 10\\ 8 & 2 \end{array}\right)\,\left(\begin{array}{cc}{y_1}(x) \\ {y_2}(x)\end{array}\right)$$
The eigenvalues are
$$\lambda_1 = 12, ~~\lambda_2 = -6$$
The eigenvectors are
$$v_1 = \begin{pmatrix} 5 \\ 4 \end{pmatrix}, ~~v_2 = \begin{pmatrix} -1 \\ 1 \end{pmatrix}$$
The solution can written as
$$y(x) = c_1 e^{\lambda_1 x} v_1 + c_2 e^{\lambda_2 x} v_2 = c_1 e^{12 x}\begin{pmatrix} 5 \\ 4 \end{pmatrix} + c_2 e^{-6x}\begin{pmatrix} -1 \\ 1 \end{pmatrix}$$
We can also use the matrix exponential
$$e^{A x} = P e^{Dx} P^{-1} = \begin{pmatrix} 5 & -1 \\ 4 & 1 \end{pmatrix}\begin{pmatrix} e^{12x} &0 \\ 0 & e^{-6x} \end{pmatrix}
\begin{pmatrix}
 \dfrac{1}{9} & \dfrac{1}{9} \\
 -\dfrac{4}{9} & \dfrac{5}{9} \\
\end{pmatrix} = \begin{pmatrix}
 \dfrac{4 e^{-6 x}}{9}+\dfrac{5 e^{12 x}}{9} & \dfrac{-5}{9} e^{-6 x}+\dfrac{5 e^{12 x}}{9} \\
 \dfrac{-4}{9} e^{-6 x}+\dfrac{4 e^{12 x}}{9} & \dfrac{5 e^{-6 x}}{9}+\dfrac{4 e^{12 x}}{9} \\
\end{pmatrix}$$
The solution using the matrix exponential is given by
$$y(x) = e^{Ax} c = \left(
\begin{array}{c}
 c_1 \left(\dfrac{4 e^{-6 x}}{9}+\dfrac{5 e^{12 x}}{9}\right)+c_2 \left(\dfrac{-5}{9}e^{-6 x}+\dfrac{5 e^{12 x}}{9}\right) \\
 c_1 \left(\dfrac{-4}{9} e^{-6 x}+\dfrac{4 e^{12 x}}{9}\right)+c_2 \left(\dfrac{5 e^{-6 x}}{9}+\dfrac{4 e^{12 x}}{9}\right) \\
\end{array}
\right)$$
Compare the two results while noting that you can combine constants because they are arbitrary.
Also, if you choose an initial condition, both methods produce exactly the same result.
