Can someone show me a proof of the general solution for 2nd order homogenous linear differential equations? I thought I had it figured out but there's a sort of 'leap of faith' at a pivotal point that annoys me.  Can someone show me how to derive the general solution of an equation such as:
$a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=0$
I want to avoid just saying, "let's assume the solution is of the form $y=e^{mx}$".  I want to see that form jumping out of an algebraic explanation. 
I know I'll get 
$y=Ae^{\alpha x}+Be^{\beta x}$
when I have 2 distinct roots, but explaining the general form for when I have two equal roots requires me to use this 'leap of faith' part.
 A: Let $u$ and $v$ be the roots of $x^2 + bx + c =0$. Then $$y''+by'+cy=0 = y''-(u+v)y'+uvy =\left(y' - uy\right)' - v\left(y' - uy\right).$$
Let $y' - uy = z (x)$. Then $$z' - vz=0  \implies z(x) = A e^{vt}.$$
Now, you have $$y' - uy = Ae^{vt}.$$ To solve this, use the integrating factor trick:
$$e^{-ux} y' - u e^{-ux}y = A e^{(v-u)x} = \left(y\cdot e^{-ux}\right)'.$$
Integrate both sides $$y\cdot e^{-ux} = \int Ae^{(v-u)x} dx = B e^{(v-u)t}+C$$
and finally get $$y(x) = B e^{vx}+Ce^{ut}.$$
Edit. As Pragabhava suggested, here's the case when $u = v$:
Suppose that we have already gotten $$A e^{(v-u)x} =\left(y e^{-ux}\right)'.$$ Then $$A e^{0} = \left(y e^{-ux}\right)' = A.$$ Integrate both sides to get $$y e^{-ux} = Ax + B\\ \implies y(x) = A e^{ux} + Bxe^{ux},$$ as desired.
A: One way to attack this is with the Exponential Shift Theorem. 

For any polynomial $P$, constant $k$ and smooth function $u(x)$,
  $P(D) (\exp(k x) u) = \exp(k x) P(D+k) u$

Here $D$ is the differentiation operator $\dfrac{d}{dx}$.  Thus e.g. for the 
polynomial $P(t) = t^2 + 2 t + 3$, $P(D) u = \dfrac{d^2 u}{dx^2} + 2 \dfrac{du}{dx} + 3 u$.
Now finding one nontrivial solution of the differential equation $P(D) y = 0$ would be easy if the constant term of the polynomial was $0$: $y=1$ would be a solution.  But $P(t+k)$ has constant term $0$ (as a polynomial in $t$) if and only if $P(k) = 0$.  Thus you look for 
roots of the polynomial.  If $P(k) = 0$, then $P(D+k) 1 = 0$, and by Exponential Shift
$P(D) \exp(k x) = 0$.  Each distinct root of the polynomial $P$ gives you a solution; it is easy to check that these are linearly independent, and if the polynomial has all distinct roots you have a fundamental set of solutions.
If the roots are not all distinct, Exponential Shift can be applied again.  Thus if
$k$ is a root of $P$ with multiplicity $m$, the terms of $P(t+k)$ in $t^j$ for $0 \le j \le m-1$ are all $0$.  Then $P(D+k) u$ involves only the $m$'th and higher derivatives
of $u$, which means that it is $0$ if $u$ is a polynomial of degree less than $m$.
So we get solutions $1, x, \ldots, x^{m-1}$ of $P(D+k) u = 0$ and by Exponential Shift solutions
$\exp(kx), x \exp(kx), \ldots, x^{m-1} \exp(kx)$ of $P(D) y = 0$.
