Repeated eigenvalue = 0 I have this quite straightforward system:
$$ \displaystyle
\begin{array}{l}
\frac{d x}{d t}=\left[\begin{array}{cc}
3 & 9 \\
-1 & -3
\end{array}\right] x \\
\text { with } x(0)=\left[\begin{array}{c}
2 \\
4
\end{array}\right] .
\end{array}
$$
My reasoning:
Using the traditional eyeball-method, one can see that this system is not linearly independant. Taking $det(A-\lambda I)$ yield $-(3+\lambda)(3-\lambda)+9 = 0 \longrightarrow \boxed{\lambda^2 = 0} $.
Since it is a repeated root, I have to get a solution in the form:
$$
\mathbf{u}(t)=c_{1} e^{\lambda t}\left(\begin{array}{l}
a_1 \\
a_2
\end{array}\right)+c_{2}\left(t e^{\lambda t}\left(\begin{array}{l}
a_1 \\
a_2
\end{array}\right)+e^{\lambda t}\left(\begin{array}{l}
b_1 \\
b_2
\end{array}\right)\right)
=e^{\lambda t}\left(c_{1}\left(\begin{array}{l}
a_1 \\
a_2
\end{array}\right)+c_{2}\left(\begin{array}{c}
b_1+a_1t \\
b_2+a_2t
\end{array}\right)\right).
$$
With eigenvectors $$
\mathbf{v_1}=\left[\begin{array}{l}
a_1 \\
a_2
\end{array}\right], 
\mathbf{v_2}=\left[\begin{array}{l}
b_1 \\
b_2
\end{array}\right], 
$$

*

*$v_1$ is then found with the null space of $A-\lambda I$ =
$$\begin{pmatrix} 3 & 9 \\ -1 & -3 \end{pmatrix} \thicksim \text{gaussa-gaussa}\thicksim \begin{pmatrix} 1 & 3 \\ 0 & 0 \end{pmatrix} \longrightarrow \left[\begin{array}{l}
a_1 \\
a_2
\end{array}\right] = \left[\begin{array}{l}
\hfill 3 \\
-1
\end{array}\right]$$


*$v2$ must fullfil the condition:
$$
\left(A-\lambda I\right) \mathbf{v}_{2}=\mathbf{v}_{1}
$$
$$\begin{pmatrix} 3 & 9 \\ -1 & -3 \end{pmatrix} \cdot\left[\begin{array}{l}
b_1 \\
b_2
\end{array}\right] = \left[\begin{array}{l}
\hfill 3 \\
-1
\end{array}\right]$$
I find that $v_2 = \left[\begin{array}{l}
\frac{1}{2} \\
\frac{1}{6} 
\end{array}\right]$ does the job.
My solution should be:
$$e^{ \lambda t}\left(c_{1}\left(\begin{array}{l}
3 \\
-1
\end{array}\right)+c_{2}\left(\begin{array}{c}
1/2+3t \\
1/6-t
\end{array}\right)\right)$$, with $e^\lambda$ disappearing since $\lambda = 0$
I find $$c_1 = -5/3 \text{ and } c_2 = 14$$
What is wrong in my reasoning?
The problem: When entering solutions, I get wrong answer:
$$
x_1 = 21t - \frac{3}{2} \\
x_2 = -7t - \frac{1}{2}
$$

Your solution is correct up until you plug the constants into your general solution:
$$-\frac53\begin{bmatrix}3 \\ -1\end{bmatrix} + 14\begin{bmatrix}\frac12 + 3t \\ \frac16 - t\end{bmatrix} = \begin{bmatrix}-5 \\ \frac53\end{bmatrix} + \begin{bmatrix}7+42t \\ \frac73 - 14t\end{bmatrix} = \begin{bmatrix}2 + 42t \\ 4 - 14t\end{bmatrix}$$
Verifying the solution:
$$x(0) = \begin{bmatrix}2 + 42(0) \\ 4 - 14(0)\end{bmatrix} = \begin{bmatrix}2 \\ 4\end{bmatrix}$$
$$x' = \begin{bmatrix}42 \\ -14\end{bmatrix}$$
$$Ax = \begin{bmatrix}3 & 9 \\ -1 & -3\end{bmatrix}\begin{bmatrix}2 + 42t \\ 4 - 14t\end{bmatrix} = \begin{bmatrix}(6 + 126t) + (36 - 126t) \\ (-2-42t) + (-12 + 42t)\end{bmatrix} = \begin{bmatrix}42 \\ -14\end{bmatrix}$$
