Sometimes a $2\times 2$ matrix has one repeated eigenvalue $\lambda$ but only one independent eigenvector $\mathbf{v}$. Then $e^{\lambda t}\mathbf{v}$ gives one solution and we need another, just as a repeated root needed $te^{rt}$. The answer looks similar, with one extra vector.
Key idea
If $\lambda$ is a double eigenvalue with only one eigenvector $\mathbf{v}$, find a generalized eigenvector $\mathbf{w}$ with $(A - \lambda I)\mathbf{w} = \mathbf{v}$. Then $$\mathbf{x} = c_1e^{\lambda t}\mathbf{v} + c_2e^{\lambda t}(t\mathbf{v} + \mathbf{w}).$$
Why it works
Try $\mathbf{x} = e^{\lambda t}(t\mathbf{v} + \mathbf{w})$. Then $\mathbf{x}' = e^{\lambda t}(\lambda t\mathbf{v} + \lambda\mathbf{w} + \mathbf{v})$, while $A\mathbf{x} = e^{\lambda t}(t\lambda\mathbf{v} + A\mathbf{w})$. These match exactly when $A\mathbf{w} = \lambda\mathbf{w} + \mathbf{v}$, that is, $(A - \lambda I)\mathbf{w} = \mathbf{v}$. Such a $\mathbf{w}$ exists because $(A - \lambda I)^{2} = 0$ for a $2\times 2$ matrix with a double eigenvalue (Cayley–Hamilton), so the column space of $A - \lambda I$ is the eigenvector line. Any $\mathbf{w}$ works; adding a multiple of $\mathbf{v}$ only changes $c_1$.
Worked example: One eigenvector
Solve $\mathbf{x}' = \begin{bmatrix} 3 & 1 \\ -1 & 1 \end{bmatrix}\mathbf{x}$, $\mathbf{x}(0) = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$.
- $\lambda^{2} - 4\lambda + 4 = (\lambda - 2)^{2}$: $\lambda = 2$ twice.
- $A - 2I = \begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix}$ has rank 1: one eigenvector, $\mathbf{v} = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$.
- $(A - 2I)\mathbf{w} = \mathbf{v}$: $w_1 + w_2 = 1$; take $\mathbf{w} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$.
- At $t = 0$: $c_1\mathbf{v} + c_2\mathbf{w} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$ gives $c_1 = 1$, $c_2 = 3$.
Answer: $\mathbf{x} = e^{2t}\begin{bmatrix} 1 + 3t \\ 2 - 3t \end{bmatrix}$
Worked example: A shortcut check
For the same $A$, write $A = 2I + N$ and compute $e^{2t}(I + tN)\mathbf{x}(0)$.
- $N = A - 2I = \begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix}$ and $N^{2} = 0$.
- $N\mathbf{x}(0) = \begin{bmatrix} 3 \\ -3 \end{bmatrix}$, so $(I + tN)\mathbf{x}(0) = \begin{bmatrix} 1 + 3t \\ 2 - 3t \end{bmatrix}$.
Answer: the same answer, through the matrix exponential
Worked example: A degenerate node
Classify the origin for $\mathbf{x}' = \begin{bmatrix} -1 & 1 \\ 0 & -1 \end{bmatrix}\mathbf{x}$.
- $\lambda = -1$ twice, with only the eigenvector $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$.
- Every solution is $e^{-t}$ times a vector that grows at most like $t$, so it decays; trajectories come in tangent to the eigenvector line.
Answer: a stable improper (degenerate) node
Common mistake
Writing $c_1e^{\lambda t}\mathbf{v} + c_2te^{\lambda t}\mathbf{v}$. That second term is not a solution: substituting it leaves an extra $e^{\lambda t}\mathbf{v}$ on the left. The correction $\mathbf{w}$ is exactly what absorbs it. Check by substituting into $\mathbf{x}' = A\mathbf{x}$.