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"?
 A: Homogeneous solutions of differential equations are solutions to the problem you stated, i.e. linear, constant coefficient equations with a zero right hand side.  One way to think of it is that you have a mathematical operator
$$
Ly = ay^{\prime\prime} + by^\prime + cy
$$ This operator takes in functions, computes the appropriate derivatives and multiplies by the constants, then spits out another function.  This operator will have zero functions, just like a quadratic equation has roots.  Those zero functions, i.e. the functions $y(t)$ such that $Ly$ is the zero function are called homogeneous solutions.  They solve 
$$
Ly = 0 
$$  
The reason why we distinguish them is that you might want to solve an inhomogeneous problem, i.e. one that looks like 
$$
Ly = f(t)
$$  This equation will have a solution that looks like $y = y_h+y_p$, where $y_h$ is a homogeneous solution (solves $Ly=0$) and $y_p$ is a solution to the full equation $Ly = f$.  If you've taken linear algebra, a homogeneous solution is the same idea as a vector in the null space of a matrix.
The reason we always use $e^{rt}$ to "find" homogeneous solutions is that exponential functions are "eigenfunctions" of the derivative.  This is just a fancy way of saying that if $y(t) = e^{rt}$, 
$$
\frac{d}{dt}y(t) = re^{rt} = ry(t)
$$  so the derivative of an exponential is a multiple of the same exponential.  This means that dropping $e^{rt}$ into $L$ will convert the differential equation into a quadratic equation for $r$!
Will a solution of the form $e^{rt}$ always work?  Yes, if you allow $r$ to be a complex number.  This is because all quadratic polynomials have 2 complex roots (counting multiplicity). 
A: 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.
