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:


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.

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    $\begingroup$ Making that «let's assume the solution is of the form...» is not to be avoided: it is to be reached! It is the result of experience. For example: when some of the terms are zero, the equation can be solved by just recognizing the solutions to be functions one is very familiar with, and all those cases can be expressed in terms of exponentials. It is natural, therefore, to try if exponentials also work in the general case. $\endgroup$ – Mariano Suárez-Álvarez Oct 14 '12 at 17:51
  • $\begingroup$ The «algebraic explanations» you look for are a somewhat guided expectation, really. Just as mathematicians do not think in terms of pure deduction and the sort of proofs that plane geometry pretends is natural, their ideas do not generally come out with «algebraic explanations»! We reason by analogy, by sheer hope, by experience, even by luck: sometimes, only sometimes, we manage to wrap the result of that reasoning in a synthetic, a posteriori «elgebraic explanation», but the result of this is quite not indicative of how we got the results. $\endgroup$ – Mariano Suárez-Álvarez Oct 14 '12 at 17:55
  • $\begingroup$ @MarianoSuárez-Alvarez +1 Fo sho! As a dilletante I am guessing thats because, most solutions can be expressed in Sin or Cos series (Fourier series) which in turn can be expressed as exponentials? $\endgroup$ – dearN Oct 14 '12 at 18:15
  • $\begingroup$ perhaps math.stackexchange.com/q/206967/36530 would be of interest to readers of this question. $\endgroup$ – James S. Cook Mar 9 '14 at 4:21

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.

  • $\begingroup$ The part you skipped (deriving the characteristic equation) is the part I need :) $\endgroup$ – Korgan Rivera Oct 14 '12 at 16:44
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    $\begingroup$ In this solution, it is seen as a trick - a trick to factorize our 2nd degree ODE to a combination of linear ones. You do not really need the characteristic equation here; making $\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=0 = \frac{d^2y}{dx^2}+(u+v)\frac{dy}{dx}+uvy =\frac{d}{dx}\left(\frac{dy}{dx} + uy\right) + v\left(\frac{dy}{dx} + uy\right)$ still works, for example. I would say that (in this answer) the roots are just a way of skipping all these steps. $\endgroup$ – Ian Mateus Oct 14 '12 at 16:51
  • $\begingroup$ The most important thing here is factoring, as I said. Splitting $b$ and $c$ into $u$ and $v$ in a way that we can work is the point here. $\endgroup$ – Ian Mateus Oct 14 '12 at 16:54
  • $\begingroup$ @IanMateus You should complement the answer with the one root case (it's right around the corner). Very nice use of factorization. (+1) $\endgroup$ – Pragabhava Oct 14 '12 at 16:57
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    $\begingroup$ @IanMateus But you did! When you solved the second first order ODE (the one for $v$), you took the integrating factor $e^{u x}$ and implicitly assumed that $u \neq v$. If $u = v$ then $$ \frac{d}{dx}\left(e^{-u x} y\right) = C$$ and voila! $\endgroup$ – Pragabhava Oct 14 '12 at 17:07

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 $P(D) y = 0$.


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