= Solution
The forcing vector has the <eigenvector> decomposition
$$
(-\lambda,1,\lambda)^T
=\frac{\lambda+1}{2}v_-+\frac{\lambda-1}{2}v_+.
$$
There is no component in the zero <eigenvalue> direction. The scalar <inhomogeneous linear differential equations> in this basis are
$$
\dot z_-=-2z_-+(\lambda+1)e^{2t},\qquad
\dot z_0=0,\qquad
\dot z_+=2z_++(\lambda-1)e^{2t}.
$$
For the first, a <particular solution> is $(\lambda+1)e^{2t}/4$. For the third, multiplying by the <integrating factor> $e^{-2t}$ gives $(e^{-2t}z_+)'=\lambda-1$. Consequently \b[for every real $\lambda$]
$$
\boxed{\begin{pmatrix}x\\y\\z\end{pmatrix}
=\left(Ae^{-2t}+\frac{\lambda+1}{4}e^{2t}\right)v_-
+Bv_0+\bigl(C+(\lambda-1)t\bigr)e^{2t}v_+.}
$$
When $\lambda=1$, the resonant component vanishes and a <particular solution> is $\tfrac12v_-e^{2t}$, of the stipulated constant-vector form. If $\lambda\ne1$, substituting $ce^{2t}$ would require $(2I-M)c=2(-\lambda,1,\lambda)^T$. In the <eigenbasis> the left side has zero $v_+$ component, while the right side has component $\lambda-1$. This is impossible. The factor $t$ is therefore unavoidable: it is <resonance in a differential equation>.
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