= Solution
At $\lambda=-1$ the $v_-$ forcing vanishes and the $v_+$ forcing coefficient is $-2$. The initial vector decomposes as
$$
(0,1,0)^T=\frac12v_- -\frac12v_+.
$$
Thus $A=1/2$, $B=0$, $C=-1/2$ in part (b), and \b[the requested solution is]
$$
\boxed{\begin{pmatrix}x\\y\\z\end{pmatrix}
=\frac12e^{-2t}v_--\left(\frac12+2t\right)e^{2t}v_+.}
$$
Equivalently,
$$
x=\frac12(1+4t)e^{2t}-\frac12e^{-2t},\quad
y=\frac12(1+4t)e^{2t}+\frac12e^{-2t},\quad
z=\frac12e^{-2t}-\frac12(1+4t)e^{2t}.
$$
These expressions have the required initial values and solve all three <linear ordinary differential equations>.
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