= Solution
Suppose $\mathbf v=\alpha\mathbf e_1+\beta\mathbf e_2$. Applying $M-\lambda_2I$ would give $\mathbf e_2=\alpha(\lambda_1-\lambda_2)\mathbf e_1$, contradicting <linear independence> of the two <eigenvectors>. Therefore the <generalized eigenvector> $\mathbf v$ lies outside their span and $(\mathbf e_1,\mathbf e_2,\mathbf v)$ is a <basis>. Write $\mathbf x=a\mathbf e_1+b\mathbf e_2+d\mathbf v$. Since $M\mathbf v=\lambda_2\mathbf v+\mathbf e_2$, the <linear system of ordinary differential equations> gives
$$
\dot a=\lambda_1a,\qquad\dot d=\lambda_2d,\qquad\dot b=\lambda_2b+d.
$$
Solving these equations gives $a=A_1e^{\lambda_1t}$, $d=A_3e^{\lambda_2t}$ and $b=(A_2+A_3t)e^{\lambda_2t}$. Thus
$$
\boxed{\mathbf x(t)=A_1e^{\lambda_1t}\mathbf e_1+e^{\lambda_2t}\left[A_2\mathbf e_2+A_3(\mathbf v+t\mathbf e_2)\right].}
$$
The <length-two Jordan chain solution> for the <Jordan chain> $(\mathbf e_2,\mathbf v)$ produces the $t e^{\lambda_2t}$ mode. The three arbitrary constants span all initial data, so this is the full general solution.
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