 Solution

For the given matrix $A$, the characteristic polynomial isFor $\lambda =4$, the eigenvector equation gives $v_2=v_1$; for $\lambda =-1$, it gives $3v_1+2v_2=0$. Thus the eigenvalues and their corresponding eigenspaces areFor any eigenvector $v$ with eigenvalue $\lambda$, the vector-valued function $e^{\lambda x}v$ satisfies $Y^{\prime }=AY$, since its derivative is $\lambda e^{\lambda x}v=e^{\lambda x}Av$. The two displayed eigenvectors are linearly independent, so the general solution of the linear system of ordinary differential equations is