Is the "homogenous solution" to a second-order linear homogenous DE always valid? That is, given an equation
$ay''+by'+cy =  0$,
I know that solutions are of the form $e^{rt},$ where r is a constant computed from $ar^2 + br + c = 0$. For some reason, I have written down in my notes adjacently that the "homogenous solution" is $c_1e^{r_1t}+c_2e^{r_2t}$. I also didn't note what precisely that was, but my assumption was that it means any equation of the form $c_1e^{r_1t}+c_2e^{r_2t}$ also satisfies the initial equation.
Is the homogenous solution really valid for $ay''+by'+cy =  0$ in all places? If so, why is the starting point that the solutions are of the form $e^{rt}$, and why is the homogenous solution valid anyway?
If not, what is meant by the "homogenous solution"?

An $n$th order, linear homogenous ODE has $n$ linearly independent solutions (See this for why.). For a $2$nd order linear homogenous ODE, we then expect two solutions.
In your case, the coefficients of the ODE are contants. This arises to the notable family of solutions that you've noted: $c_1e^{r_1t}+c_2e^{r_2t}$.
You've likely seen this, but let's assume $y=e^{rt}$. Upon substitution into the ODE, we can see our assumption is satisfied and indeed this is one of many solutions:
$$ay''+by'+cy =  0$$
$$a(e^{rt})''+b(e^{rt})'+ce^{rt} =  0$$
$$ar^2e^{rt}+bre^{rt}+ce^{rt} =  0$$
$$ar^2+br+c =  0$$
Note that the division by $e^{rt}$ is permitted since $e^{rt}>0$. This last line, though, is just a quadratic equation with roots:
$$r = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$$
The two roots give $r_1$ and $r_2$, and thus let us construct a general solution. Since it is linear, we sum the two solutions (which happen to be linearly independent. Look up Wronksian for further info). The summation of solutions is a solution itself, hence 
$$y = c_1e^{r_1t}+c_2e^{r_2t}$$
The above is your real-valued solution for DISTINCT roots $r_1 \ne r_2$. 
But what if $r_1 = r_2$ in the case of repeated roots? Or $r_1, r_2 \in \mathbb{C}$ in the case of complex roots? There are workarounds for these cases, and the answer will be again represented as a sum of two linearly independent solutions.
