$e^{mx}$ in solving second order differential equations In a book I am reading on differential equations, the author writes about the solution to a homogenous, linear, second order differential equation with constant coefficients. The author says something like, "Let us suppose that the solution is of the form $y=e^{mx}$ " . After this the author introduces the characteristic equation of a differential equation of the form mentioned above, and proceeds to describe how to solve it w/ undetermined coefficients, variation of parameters, etc. 
How did mathematicians first come up with this "assumption" that the solution was of the form $y=e^{mx}$? And, are there any other forms of solutions for these types of equations?
 A: I like to think of it as an extension of the solution of the differential equation
$$
\frac{\mathrm{d}y}{\mathrm{d}x}=ky
$$
which, as can be easily checked using separation of variables, is
$$
y=y_0\mathrm{e}^{kx}.
$$
Now if we have a second order ODE
$$
\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+a\frac{\mathrm{d}y}{\mathrm{d}x}+by=0
$$
where $a,b\in\mathbb{R}$, it is possible to re-write this as a pair of first-order ODEs. If we set $z=\frac{\mathrm{d}y}{\mathrm{d}x}$ (one of our equations) then we have the system of coupled ODEs
$$\cases{\dfrac{\mathrm{d}y}{\mathrm{d}x}=z \\\dfrac{\mathrm{d}z}{\mathrm{d}x}=-by-az}
$$
and this can be solved by solving the equation $\dfrac{\mathrm{d}\boldsymbol{y}}{\mathrm{d}x}=A\boldsymbol{y}$, where $\boldsymbol{y}=(y(x),z(x))^\top$ and $$A=\left(\begin{matrix}0 & 1\\
-b & -a
\end{matrix}\right)$$
and since this is a linear ODE (in more than one variable), it can be solved in an analogous way (using an integrating factor method)
\begin{aligned}\dfrac{{\rm d}\boldsymbol{y}}{{\rm d}x}-A\boldsymbol{y} & =0,\\
{\rm e}^{-\int A{\rm d}x}\dfrac{{\rm d}\boldsymbol{y}}{{\rm d}x}-A\boldsymbol{y}{\rm e}^{-\int A{\rm d}x} & =0,\\
\dfrac{{\rm d}}{{\rm d}x}\left(\boldsymbol{y}{\rm e}^{-\int A{\rm d}x}\right) & =0,\\
\dfrac{{\rm d}}{{\rm d}x}\left(\boldsymbol{y}{\rm e}^{-Ax}\right) & =0,\\
\boldsymbol{y}{\rm e}^{-Ax} & =\boldsymbol{C},\\
\therefore\boldsymbol{y}(x) & =\boldsymbol{C}{\rm e}^{Ax}.
\end{aligned}
I'm sure with some more effort, it will be possible to recover linear combinations of exponentials in the individual components of the solution. Probably seems a bit rambly, but this is one way I like to convince myself where the exponential solutions come from - it is the solution to an aforementioned first-order differential equation, so by linearity, more exponentials will appear in the solution. (Don't know if that explains things well enough.)
A: Probably the idea comes from the first order equation, where
$$
y'+ky=0
$$
could be solved by separation of variables
$$
\frac{y'}{y}=-k\implies \ln y=-kx+c\implies y=Ce^{-kx}.
$$
A: $e^{mx}$ turns out to be the general form for solutions to linear differential equations with constant coefficients.  One can also show that solutions to "equipotential" equations, equations of the form $ax^2y''+ bxy'+ cy= 0$ tend to be $x^n$ for some number $n$.
A: This captures previous comments in a slightly different formalism. 
Taking a derivative is a linear operator:Given functions $f$ and $g$, and real constants $\alpha$ and $\beta$, $$d(\alpha f + \beta g)/dx=\alpha df/dx+\beta dg/dx$$
If a linear operator returns a scalar multiple of its input, the input is an Eigen Vector and the scalar multiple is an Eigen Value. 
Given $$d(e^{mx})/dx=me^{mx}$$, $m$ is the Eigen Value and $e^{mx}$ is the Eigen Vector, or Eigen Function in this case. 
This allows for various concepts of linear algebra to be applied to the solutions of these problems. 
