Solution of system of linearly dependent equations. So, I have the system of equations $x'(t) = Ax$ where $A$ is first row-(4,-2) and second row - (8,-4). This has two eigenvalues, both are 0. But I tried to solve it this way:
$x_1' = 4x_1 -2x_2$

and

$x'_2 = 8x_1 -4x_2$


Next, I multiplied the top equation by 2 and got $2x_1' = 8x_1 - 4x_2$
Then I equated $2x_1'$ and $x_2'$ so $2x_1' = x_2'$ and integrating both sides I have $2x_1 = x_2$. 
So, it seems to me that the solution should just be $x_1 = f(x)$ and $x_2 = 2f(x)$ where $f(x)$ is any arbitrary function. But apparently that isn't valid. I'm sure I'm missing something huge/obvious but I'm lost as to why that doesn't work.
 A: We have the system:
$$x'(t) = Ax = \begin{bmatrix}4 & -2\\8 & -4\end{bmatrix}$$
If we solve for the characteristic polynomial, we have:
$$|A - \lambda
 I| = 0 \rightarrow \lambda^2 = 0 \rightarrow \lambda_{1,2} = 0$$
Now, we take a eigenvalue and use it to solve:
$$[A - \lambda_1 I]v_1 = 0$$
We then need a generalized eigenvector and use:
$$[A - \lambda_1 I]v_2 = v_1$$
We should get:
$$\displaystyle v_1 = (1, 2), v_2 = \left(\frac{1}{4},0\right)$$
A: Hint:
Set $x_1=x$ and $x_2=y$, so your system would be: $$
\left\{
        \begin{array}{ll}
            x'=4x-2y \\
            y'=8x-4y
        \end{array}
    \right.
~~~~~\text{or}~~~~~
\left\{
        \begin{array}{ll}
            Dx=4x-2y \\
            Dy=8x-4y
        \end{array}
    \right.
$$ We can be go further: $$\left\{
        \begin{array}{ll}
            (D-4)x+2y=0 \\
            (D+4)y-8x=0
        \end{array}
    \right.$$ Trying to solve this system as an ordinary system we get: $$\left\{
        \begin{array}{ll}
            D^2x=0 \\
            D^2y=0
        \end{array}
    \right.$$
