Substituting the proposed form $\mathbf v(t)=e^{\lambda t}\mathbf u$ into the <linear system of ordinary differential equations> gives
$$
e^{\lambda t}(\lambda I-B)\mathbf u=e^{\lambda t}\mathbf x.
$$
If $\lambda$ is not an <eigenvalue> of $B$, the matrix $\lambda I-B$ is <invertible matrix>[invertible], so
$$
\mathbf u=(\lambda I-B)^{-1}\mathbf x
$$
provides the required particular solution.

In the stated two-dimensional case,
$$
2I-B=\begin{pmatrix}2&-3\\-1&2\end{pmatrix},
\qquad
(2I-B)^{-1}=\begin{pmatrix}2&3\\1&2\end{pmatrix},
$$
and therefore $\mathbf u=(3,2)^T$. The <eigenvalues> of $B$ are $\pm\sqrt3$, with corresponding <eigenvectors> $(\sqrt3,1)^T$ and $(-\sqrt3,1)^T$. Adding the <general solution> of the homogeneous system yields
$$
\boxed{
\mathbf v(t)=e^{2t}\begin{pmatrix}3\\2\end{pmatrix}
+C_+e^{\sqrt3t}\begin{pmatrix}\sqrt3\\1\end{pmatrix}
+C_-e^{-\sqrt3t}\begin{pmatrix}-\sqrt3\\1\end{pmatrix}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 2 / 2C / Solution

The general solution of a <differential equation> is a family containing every solution, usually parameterized by arbitrary constants or initial data. For a forced <linear ordinary differential equation>, it is the sum of one <particular solution> and the general <homogeneous solution>.

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