 Solution

Substituting the proposed form $\mathbf{v}(t)=e^{\lambda t}\mathbf{u}$ into the linear system of ordinary differential equations givesIf $\lambda$ is not an eigenvalue of $B$, the matrix $\lambda I-B$ is invertible, soprovides the required particular solution.

In the stated two-dimensional case,and therefore $\mathbf{u}=(3,2{)}^T$. The eigenvalues of $B$ are $\pm \sqrt{3}$, with corresponding eigenvectors $(\sqrt{3},1{)}^T$ and $(-\sqrt{3},1{)}^T$. Adding the general solution of the homogeneous system yields