Where does the guessed solution come from when solving DE systems with a double root?

Say I have

$\vec{v}' = \begin{bmatrix}-2 &1\\-1 & 0\end{bmatrix}\vec{v}$

Solving this up to the characteristic equation, I get a double root $\lambda=-1$.

Assuming I didn't know any better, I'd move forward to find a solution

$\begin{bmatrix} 1 \\ 1 \end{bmatrix}e^{-t}$

At this point, all the explanations I've found so far online have said "let's then try using the solution multiplied by t to see if that works" since that's the approach used when solving a 2nd-order equation. For this problem, it will fail. Then the explanation usually goes "Then, let's try

$$ \vec{v}=t\vec{n}e^{-t}+\vec{p}e^{-t} $$

and everything works"; you end up with terms

$$ (A-\lambda I)\vec{n}=0 \\ (A-\lambda I)\vec{p}=\vec{n} $$

and solve from there.

Let's say I just happened to know the first guess where the form of a solution is $\vec{v}=e^{\lambda t}p$. How would I then figure out from scratch that I could then add on $te^{\lambda t}\vec{n}$ to get everything to fall into place?

The system is equivalent to $$x'=-2x+y$$

$$y'=-x$$

solving for $x$ provides $$x''+2x'+x=0$$ with characteristic root of $-1$ with multiplicity of $2$

One solution is $$x=e^{-t}$$

For the other solution we let $$x=u(t)e^{-t}$$ and solve for $u(t)$

Plug in $$x''+2x'+x=0$$ and you come up with u(t)=t, hence the second solution is $$ x=te^{-t}$$ and the general solution $$ x=(C_1 + C_2 t) e^{-t}$$

We can find $y$ from $$y=x'+2x$$