My problem is as follows:

Given the second order homogeneous linear differential equation with constant coefficients $$a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+c\,y(x)=0,$$ is there a good heuristic that explains why the solution set is of the form $\{Ae^{\alpha x}+Be^{\beta x}\}$ or $\{Ae^{\alpha x}+Bxe^{\alpha x}\}$.

The background is that I am teaching engineers the method of solving these equations but like everything else I like to give them a reason why the method works.

I can explain why we might look for solutions of the form $e^{rx}$, why something like $Ae^{rx}$ will be a solution and why if $y_1$ and $y_2$ are solutions then so is $y_1+y_2$. I can explain the non-homogeneous case and why occasionally we have to look at test solutions of the form $xy_H$ --- where $y_H$ is a solution of the homogeneous equation.

The problem occurs when I try and explain to them why the solutions have to be two dimensional and that we don't need three linearly independent solutions (in the homogeneous case).

My best hand-waving argument thus far is that in a solution we will have to integrate twice somewhere and so we will end up with two constants of integration say $C_1$ and $C_2$ so our solution will be $$y_H=y(x,C_1,C_2)$$ but I have had to wave very hard indeed to turn this into $y_H=Ay_1+By_2$.

These are not maths students but I still tried to make various bad arguments along the lines of the 'kernel' of the operator $\displaystyle D^2=\frac{d^2y}{dx^2}$ being two dimensional and that the addition of $bD$ and $cI$ distorts the 'kernel' but not the dimension of it (I wonder can this argument be made rigorous).

Have any of ye any better ideas? I understand that we can show from the Uniqueness and Existence Equation that the solutions must have this form... the irony is that I am happy to sketch an argument of plausibility of that fact --- which is left without proof in most ODE classes --- but the journey from there to the conclusion, which is done in these classes, is beyond the scope and interest of this class.

I fully expect a comment along the lines of they're engineers, who cares?

I do!

  • 2
    $\begingroup$ I'm not an engineer, but the existence+uniqueness of a solution with a given $y(0)$ and $y'(0)$ is IMHO essential for practical understanding of what the ODE does: I give you initial position and velocity, and the equation makes you move by specifying the acceleration (changing it to a system of 1st order equations might make it more intuitive). They are going to solve problems with initial conditions, so they need to see that for every initial condition there is unique $A$ and $B$. (this is not a "mathematical", but rather "intuitive" comment) $\endgroup$
    – user8268
    Sep 19, 2013 at 21:26
  • 2
    $\begingroup$ Rebel engineers only care about one thing. Or maybe two, if its happy hour. $\endgroup$
    – copper.hat
    Sep 19, 2013 at 22:12
  • $\begingroup$ I added an answer below. I graduated from UCC. Elec. Eng. in 1982, and was taught by some of the early greats like George Kelly, Michael O'Callaghan (a favourite) and Finbarr Holland. $\endgroup$
    – copper.hat
    Sep 20, 2013 at 6:00

5 Answers 5


I don't mean to put anyone down, but everything above will invariably prove too "mathematicky" for some of the engineers. (No wonder, as it refuses to leave its math nest even for a second.) Explaining that $y=C_1 y_1 + C_2 y_2$ can be done simpler & (thankfully) less rigorously by means of the following two arguments:

First, your engineers will recognize this as Newton's law, which they're familiar with exactly b/c they're engineers. It's not all that hard to convince them that you need to give your general solution "space" to accommodate... what do we need to accommodate (ask)? What does Newton require to give us the full trajectory of a body - e.g., of a body attached to a spring & a damper (as in your example)? Pat verbally on the back the first person to say 'position & velocity.' That's two constants - great. But how are these constants combined in the general solution?

Well, let's take a look at that ODE again. How would I solve it if I didn't know that I have to look for exponentials & all? What I'd do is, I'd take my ODE & initial position/velocity (write these down) & solve the ODE for $y''$ (solve it). I get $y'' = (-b/a)y' - (c/a)y$, which is a linear combination. I can do the same trick w/ the 3rd derivative by differentiating (both sides of) the eq. (do it explicitly) & replacing the 2nd derivative in it by the lin. comb. of the first two - voila (show how it's done), another linear combination. & so on - they should now be happy to believe you. Why is that useful? But, of course, because I have Taylor expansions in my mind. Write down Taylor up to $t^3-$terms & plug in for the coefficients. Rearrange into something of the form $C_1 y_1(t) + C_2 y_2(t)$ (really, really do that). There you go.

Btw, anyone who uses engineering products should care exactly b/c they are engineers, & that would be... wait, just about everybody, incl. people who ride cars & trains & planes moving at potentially crushing - for them - speeds, people living close to nuclear/chemical factories or bombs (who doesn't?), & so on. But it's not as if math zealots will disappear overnight...

  • $\begingroup$ Excellent thank you. The other answers are all superb but beyond the mathematical sophistication of the class (at the moment). $\endgroup$ Sep 20, 2013 at 12:50
  • $\begingroup$ @JpMcCarthy: What year engineering are you teaching? $\endgroup$
    – copper.hat
    Sep 20, 2013 at 20:59
  • $\begingroup$ @copper.hat They are in third year of an ordinary degree out in what was the RTC. $\endgroup$ Sep 21, 2013 at 16:02
  • $\begingroup$ @JpMcCarthy: Thanks! Their maths. syllabus must be quite a bit different than the UCC one was in 'my day'. $\endgroup$
    – copper.hat
    Sep 21, 2013 at 16:14

For the model problem $$\frac{d^2y}{dt^2}-b\frac{dy}{dt}-c\,y(t)=0$$ let $x_1=y(t)$, $x_2=y'(t)$ and write as a system: \begin{align} x_1'(t) &= x_2(t) \\ x_2'(t) &= c\,x_1(t) + b\,x_2(t). \end{align} Letting $\mathbf{x}=\begin{bmatrix} x_1 & x_2 \end{bmatrix}^T$ write in matrix form as $$ \mathbf{x}'(t) = \begin{bmatrix} 0 & 1 \\ c & b \end{bmatrix} \mathbf{x}(t). $$ Now you just need to believe in/motivate exponential solutions for the first order system $\mathbf{x}'=A\mathbf{x}$. It's easy enough to see that if $\mathbf{v}$ is an eigenvector for $A$ with eigenvalue $\lambda$ that $e^{\lambda t}\mathbf{v}$ is a solution. Now hopefully their linear algebra knowledge kicks in, making the nature of the dimension of the solution space very clear.

The matrix form is supremely important for engineers. It's helpful for visualizing stability analysis, it explains the word characteristic polynomial, it motivates diagonalization and Jordan normal form of matrices, it has connections to solving PDEs numerically... I could go on... but most importantly it should generate questions in the student's mind:

  • What happens if we have fewer eigenvectors than the order of the equation?
  • What happens with repeated eigenvalues?
  • Does this work for higher order equations?
  • What about nonhomogeneous equations?
  • $\begingroup$ Yep, this is something which I had in mind when I read the question. Moreover, as soon as the matrix exponent is defined, the uniqueness theorem can be proved in just one line and without any hand waving. +1. $\endgroup$
    – Artem
    Sep 20, 2013 at 2:01

I hope to be helpful.

1- If $\alpha\neq \beta$, for any initial condition $y(t_0)=y_0$ and $y'(t_0)=y_1$ we always can solve $$ Ae^{\alpha t_0}+Be^{\beta t_0}=y_0 \\ A\alpha e^{\alpha t_0}+B\beta e^{\beta t_0}=y_1 $$ since $$ \left(\begin{array}{cc}e^{\alpha t_0} & e^{\beta t_0} \\ \alpha e^{\alpha t_0} & \beta e^{\beta t_0}\end{array}\right) $$ is no singular.

Then there is no need of a third function to solve the given differential equation.

Same idea if $\alpha=\beta$.

2- Factorization: $$ ay''+by'+cy=a\,(\frac{d }{d x}-\alpha I)( \frac{d}{dx}-\beta I)\,[y] $$

  • $\begingroup$ I kind of like the first reason in that it says you definitely need two which I didn't hit on before. I could then say to engineers that if two is enough you don't need three. $\endgroup$ Sep 20, 2013 at 6:22

I think it is essential that engineers understand that under certain conditions, the ODE has a unique solution for every initial condition, but you are looking for intuition here. You can deal with this without appealing to contractions and completeness of $C^n[0,1]$. Whether or not what follows is intuitive is arguable, but there is no hand waving.

To simplify life, let $D = \frac{d}{dx}$, as usual. First, consider the equation $(D-\lambda) x = f$, $x(0) = x_0$, where $\lambda \in \mathbb{C}$, and $f$ is continuous. We are looking for $C^1$ solutions on, say, $[0,1]$. Consider the following steps: $$ (D-\lambda) x = f $$ $$ e^{-\lambda t} ((D-\lambda) x)(t) = e^{-\lambda t} f(t) $$ $$ D( s \mapsto e^{-\lambda s} x(s) ) (t) = e^{-\lambda t} f(t) $$ $$ e^{-\lambda t} x(t) - x_0 = \int_0^t e^{-\lambda \tau} f( \tau ) d \tau$$ $$ x(t)= e^{\lambda t } x_0 + \int_0^t e^{\lambda (t-\tau)} f( \tau ) d \tau$$ There are two points here: (1) The last equation gives an explicit solution, so existence is given. (2) If $\tilde{x}$ is another solution, then the above steps show that $\tilde{x} = x$, hence we have uniqueness. Setting $f=0$ shows why the equation $(D-\lambda) x = 0$ has solutions of the form $t \mapsto e^{\lambda t } x_0$, and why the entire solution depends on a single initial condition $x_0$. This sets the stage for the 2nd order equations.

Now consider equations of the form $(D-\lambda_1) (D-\lambda_2) x = 0$, $x(0) = x_0$, $x'(0) = x_0'$, that is, any linear time-invariant (LTI) 2nd order system. We split this into two systems: (1) $(D-\lambda_1) g = 0$, $g(0) = g_0 = x_0' - \lambda_2 x_0$, and (2) $(D-\lambda_2) x = g$, $x(0) = x_0$.

We see immediately that $ g(t)= e^{\lambda_1 t } g_0 $, and $ x(t)= e^{\lambda_2 t } x_0 + \int_0^t e^{\lambda_2 (t-\tau)} e^{\lambda_1 \tau } g_0 d \tau$. The result of the integral depends on whether or not $\lambda_1 = \lambda_2$. In either case, it is immediate that the (as in 'the only') solution is in the span of $t \mapsto e^{\lambda_k t}$ (or $t \mapsto e^{\lambda_1 t}$ and $t \mapsto t e^{\lambda_1 t}$, if $\lambda_1 = \lambda_2$).

The same reasoning as above gives existence and uniqueness.

The above shows that the solution depends on two dimensions of initial conditions.

As an aside, I think the above leads very naturally into the matrix form (that is, a first order system in two variables), relationship to eigenvalues, Jordan form, higher order systems, exponential of a matrix, etc. It would be appropriate to point out that much of this generalises beyond finite-dimensional LTI systems, but more machinery and assumptions are necessary.


Existence and uniqueness

Let's consider \begin{align*} \frac{d}{dt}\bigg[\frac{d}{dt}\bigg(x(t)\,e^{at}\bigg)\,e^{bt}\bigg] =0&&(1) \end{align*} Then \begin{align*} \frac{d}{d\tau}\bigg(x(\tau)\,e^{a\tau}\bigg)\,e^{b\tau}\bigg|_{t_0}^t &=0\\ \frac{d}{dt}\bigg(x(t)\,e^{at}\bigg) &=\dot x(t_0)\,e^{-bt+(a+b)t_0}+a\,x(t_0)\,e^{-bt+(a+b)t_0} \end{align*} In order to proceed with further integration, we identify two cases

  • Case $b=0$ \begin{align*} \frac{d}{dt}\bigg(x(t)\,e^{at}\bigg) &=\dot x(t_0)\,e^{a t_0}+a\,x(t_0)\,e^{a t_0}\\ x(t)\,e^{at} &=x(t_0)\,e^{at_0}+\dot x(t_0)\,e^{a t_0}\,(t-t_0)+a\,x(t_0)\,e^{a t_0}\,(t-t_0)\\ x(t) &=x(t_0)\,e^{-a(t-t_0)}+\dot x(t_0)\,e^{-a(t-t_0)}\,(t-t_0)+a\,x(t_0)\,e^{-a (t-t_0)}\,(t-t_0) \end{align*} Eventually, solution for this case is of the form $$\boxed{x(t)=(c_1+c_2\,t)\,e^{-at}}$$
  • Case $b\neq 0$ \begin{align*} \frac{d}{dt}\bigg(x(t)\,e^{at}\bigg) &=\dot x(t_0)\,e^{-bt+(a+b)t_0}+a\,x(t_0)\,e^{-bt+(a+b)t_0}\\ x(t)\,e^{at} &=x(t_0)\,e^{at_0}+\dot x(t_0)\,e^{-bt+(a+b)t_0}+a\,x(t_0)\,e^{-bt+(a+b)t_0}\\ x(t) &=x(t_0)\,e^{-a(t-t_0)}-{1\over b}\,\dot x(t_0)\,e^{-a(t-t_0)-b(t-t_0)}-{a\over b}\,x(t_0)\,e^{-a(t-t_0)-b(t-t_0)} \end{align*} Eventually, solution for this case is of the form $$\boxed{x(t)=c_1\,e^{-at}+c_2\,e^{-(a+b)t}}$$

With a few calculations, it's possible to prove that for suitable $a,b$ the original problem can be rewritten as problem (1), so to arrive to its solution in a direct way, and so to prove existence and uniqueness of the solutions found.

Substitutions required are the following \begin{align*} \left\{\begin{matrix} a_{1,2}={\bar b \pm \sqrt{\bar b^2-4\,\bar c\bar a}\over 2\bar a}\\ b_{1,2}=\mp {\sqrt{\bar b^2-4\,\bar c\bar a}\over \bar a} \end{matrix}\right. \end{align*} where $\bar a,\bar b, \bar c$ are the coefficients of the original problem.


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