= Solution
Substituting an exponential particular solution gives $\lambda u-Au=v$, or $(\lambda I-A)u=v$. Since $\lambda$ is not an <eigenvalue> of $A$, the matrix is invertible. The <exponential forcing of a constant-coefficient differential system> therefore has
$$
\boxed{x_p(t)=e^{\lambda t}(\lambda I-A)^{-1}v.}
$$
For the specified matrix, $I-A=\begin{pmatrix}1&-1\\0&1\end{pmatrix}$. Solving $u_2=1$, $u_1-u_2=1$ gives $u=(2,1)^T$, so
$$
\boxed{x_p(t)=e^t\begin{pmatrix}2\\1\end{pmatrix}.}
$$
Differentiation verifies both component equations directly. If a general solution were desired, the homogeneous term $e^{tA}c$ would be added, with the <matrix exponential> here equal to $I+tA$ because $A^2=0$.
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