By theorem 6.19 we know that the solution is

\(\displaystyle{y}={c}_{1}{\left({e}^{{λ}{1}{t}}\right)}{u}_{1}+\ldots+{c}_{n}{\left({e}^{{λ}_{n}{t}}\right)}{u}_{n}\)

with \(\lambda_i\) the eigenvalues of the matrix A and \(u_i\), the eigenvalues. Thus for this case we then obtain the general solution:

\(\left[\begin{array}{c}y_1\\ y_2\end{array}\right]=y=c_1e^t\left[\begin{array}{c}2\\ -1\end{array}\right]+c_2e^3t\left[\begin{array}{c}3\\ 1\end{array}\right]\\\)

Thus we obtain: \(y_1=2c_1e^t+3c_2e^3t\)

\(y_2=-c_1e^t+c_2e^3t\)