In a system of ODEs with a repeated root, what brings one to assume that the form of the equation is
\begin{align} y(t)=v(t)e^{rt} \end{align}
where $r$ is the repeated root. I get that multiplying $ce^{rt}$ by $t$ yields a second, linearly independent solution, but how would you arrive at the above supposition intuitively?
In ODE and PDE research, sometimes a "guess and check" method is employed. Multiplying by $ce^{rt}$ works nicely.
Consider the ODE $y'=y$ with initial condition $y(0)=k$. Without knowing the solution is the exponential, we want to find a function (more generally, a family of functions) that is equal to its own derivative. There is only one such function that is equal to its own derivative everywhere (not just almost everywhere).
This is perhaps remarkable, because it is essentially saying out of the uncountably many functions in $C^1$, only one of them is equal to its own derivative. The proof of this fact however requires some justification, and I remember correctly is not an entirely trivial proof.
For a system of ODEs, the solution $y(t)=v(t)e^{rt}$ works. We can check to see that is unique (under some initial conditions). However, a priori, without knowing the solution to the system takes this form, we would have to use a lot of machinery to rule out other possible functions as solutions.