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?
Best Answer
There are two expressions $$ \tag{1} y(x) = \exp(Ax) C, $$ and $$ \tag{2} y(x) = c_1 v_1\exp(\lambda _1 x) + \cdots + c_n v_n \exp(\lambda _n x), $$ where $C = \begin{pmatrix} C_1 \\ \vdots \\ C_n\end{pmatrix}$ and we write $c = \begin{pmatrix} c_1 \\ \vdots \\ c_n\end{pmatrix}$.
Since we know $\exp(Ax) = V \exp(\Lambda x) V^{-1}$,
\begin{align} \exp(Ax) C &= V \exp(\Lambda x) V^{-1} C \\ &=\begin{pmatrix} v_1 & \cdots & v_n\end{pmatrix} \begin{pmatrix} \exp(\lambda_1 x) & & \\ & \ddots & \\ & & \exp(\lambda_n x) \end{pmatrix} V^{-1}C \\ &= \begin{pmatrix} \exp(\lambda_1 x) v_1 & \cdots & \exp(\lambda_n x)v_n\end{pmatrix} V^{-1}C \end{align}
since (2) can be written as $ \begin{pmatrix} \exp(\lambda_1 x) v_1 & \cdots & \exp(\lambda_n x)v_n\end{pmatrix} c$, we see that $C$ and $c$ has a simple relation $$ C = Vc.$$