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?

  • $\begingroup$ Well, exponentials are good first guess. And it turns out they work, together with the appropriate treatment for repeated roots. Then there are theorems that guarantee that once we have found such solutions there are no other solutions. Keep studying ODE theory & real analysis and you'll eventually get to those results. $\endgroup$
    – Simon S
    May 30 '15 at 20:18
  • $\begingroup$ This is a good question, and it's always good to try to figure out how people came up with ideas. But I do not think there is a satisfying answer other than "Someone thought to try it, and it works every time." This wouldn't be the first or the last time that problems get solved in that way. $\endgroup$
    – davidlowryduda
    May 30 '15 at 20:21
  • $\begingroup$ It seems if you ask a question like this here, you get guesses. Maybe to get actual references on "how did they first come up with that?", you should ask in hsm.stackexchange.com . $\endgroup$
    – GEdgar
    May 31 '15 at 0:07
  • $\begingroup$ Like a lot of things in mathematics, you start with a good guess, and then prove that guess. $\endgroup$
    – K.defaoite
    Jun 1 '21 at 14:12

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.)

  • $\begingroup$ There is a wrong sign between the first and second equation. $\endgroup$
    – enzotib
    May 31 '15 at 13:25
  • $\begingroup$ Sorted now, cheers! $\endgroup$
    – user240560
    May 31 '15 at 19:56

Probably the idea comes from the first order equation, where $$ y'+ky=0 $$ could be solved by separation of variables $$ \frac{y'}{y}=-k\implies \ln y=-kx+c\implies y=Ce^{-kx}. $$


$e^{mx}$ 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 $n$.

  • $\begingroup$ The TeX delimiter are $ for inline equation, $$ for centered equation. ** to delimit bold. $\endgroup$
    – enzotib
    Jun 1 '15 at 6:39

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.


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