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This question might take a while. Bear with me.

I'm going to explain how I solve these up to the point where this textbook does something unusual, then I'll explain why it's unusual to me and ask for advice.

Let's say I have:


I'd start with assuming $a=0$:


Then I rearrange:


The general solution of this is:


Let's assume that b and c are known constants.

Then I figure out the 2nd and 3rd derivatives:

$y' = -\frac{kc}{b}e^{-\frac{c}{b}x}$

$y'' = \frac{kc^2}{b^2}e^{-\frac{c}{b}x}$

Now, the textbook I'm using does this. Let's say that:


So the derivatives above become


$y' = kme^{mx}$

$y'' = km^2e^{mx}$

I then substitute this into my original equation to get:

$akm^2e^{mx} +bkme^{mx}+cke^{mx}=0$


$ke^{mx}(am^2 +bm+c)=0$

The idea, now is that since $ke^{mx}$ can't be zero, I solve this quadratic in $m$ and proceed from there.

However, this is my sticking point. $m=-\frac{c}{b}$ So

$ke^{mx}(am^2 -c +c)=0$


Since $ke^{mx}$ again can't be zero, I really have to solve






At this point, I don't know if I'm alone in forest having wandered off the path hours ago. :)

Look, guys: I KNOW that assuming that a is zero reduced this to a first order DE. I do this, as I have already explained, to find y in terms of b and c. Then, I take this new y and substitute into my original equation WHERE a IS NOT ZERO.

MY QUESTION IS, if I have to solve a quadratic in m, but the quadratic falls apart, then what? (To reiterate, a in the quadratic is NOT zero.)

later edit: As far as I can tell, saying that -c/b =m seems to be a clever way to allow the quadratic to exist and then move onto finding roots, thereby allowing me to write a general solution in terms of those roots. So that's my answer. At least, it's the best I have found so far.

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You already assumed that $a=0$, so the last term,$am^2$, is zero no matter what $m$ is. – Daryl Sep 20 '12 at 20:09
$a$ was allowed to be zero in the beginning so I could find $y$ in terms of $b$ and $c$. After I substitute this back into the original equation, $a$ being zero is no longer required and is assumed to be a known constant again. The problem is that I'm solving a quadratic in m, but my quadratic falls apart. – Korgan Rivera Sep 20 '12 at 20:14
The problem requires a completely different solution style for $a=0$ and $a\neq 0$. When $a=0$, the ODE is only first order. When $a\neq0$, the ODE is second order. You can't use the same techniques, in general, for solving both of these. – Daryl Sep 20 '12 at 20:20
Like I already said, setting $a$ to zero temporarily allows me to find y in terms of $b$ and $c$. I substitute this result into the ORIGINAL equation where $a$ is NOT ZERO. So it remains a 2nd order DE. The question is not about $a$, it's about $m$. – Korgan Rivera Sep 20 '12 at 20:47
Then that book gives very bad, potentially wrong, advice for solution techniques of higher order ODEs. – Daryl Sep 20 '12 at 20:55
up vote 1 down vote accepted

When $a=0$, the ODE is first order and a myriad of techniques are available to solve this.

When $a\neq 0$, the ODE is second order. To solve higher order ODEs, with constant coefficients, assume a 'trial' solution of $$y=e^{mx}.$$ If this satisfies the original ODE, we need $$\left(am^2+bm+c\right)e^{mx}=0.$$ As you argued above, $e^{mx}\neq0$, so we need the quadratic to equal zero. To solve the quadratic, look up quadratic formula.

It is these values for $m$ which solve the problem when $a\neq0$. They are (nearly) unrelated to the solution when $a=0$.

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Look at the quadratic. Replace m with -b/c. Simplify the quadratic. See my problem? – Korgan Rivera Sep 20 '12 at 20:42
But when $a\neq0$, $m$ is not $-b/c$. $m$, by definition, MUST solve this equation and you get two entirely different forms of solution for the two cases $a=0$ and $a\neq0$. For starters, since the equation is quadratic, you will get 2 (possibly equal) values for $m$. You only have 1 because you solved the wrong equation for $m$. You need to solve the one I gave to get solutions when $a\neq0$. – Daryl Sep 20 '12 at 20:52
m is defined as -b/c. It's just for convenience. Really I'm solving a quadratic with respect to -b/c. But if the second term in the quadratic can be reduced to -c, then the thing falls apart. I'm going to try to figure this out and I'll be back. – Korgan Rivera Sep 20 '12 at 20:58
I suggest abandoning your book (it sounds wrong) and looking up solution techniques for constant coefficient higher order ODEs on the web. – Daryl Sep 20 '12 at 21:01
What you said is wrong. Should I abandon you? – Korgan Rivera Sep 20 '12 at 22:12

Perhaps you're misinterpreting what the book is trying to say. The point is that an exponential solves the differential equation in the case $a=0$, so we try exponentials (but not the same one!) when looking for a solution in the case $a \ne 0$.

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No, I'm not misinterpreting it. This is the technique it uses to first find y in terms of b and c, before reintroducing it to the original equation where a is not zero. – Korgan Rivera Sep 20 '12 at 20:42
Who knows. You're right: m is a constant but I treat it like a variable. It works but I don't know why. – Korgan Rivera Sep 23 '12 at 17:50

Here is the solution!

So, while it is true that the quadratic does in fact collapse, given that m=-c/b, the trick is to not allow it to do so. You can solve the quadratic regardless, with respect to -c/b (or m if you prefer), for two roots, say $\alpha$ and $\beta$. Here's how I do it:


set $a = 0$ (TEMPORARILY!)


solve this for y.

$y = ke^{-\frac{c}{b}x}$

where k is an unknown constant. Find the 1st and second derivatives of y.

$y' = -\frac{c}{b}ke^{-\frac{c}{b}x}$

$y'' = (\frac{c}{b})^2ke^{-\frac{c}{b}x}$

Substitute this into the original equation



$ke^{-\frac{c}{b}x} (a(\frac{c}{b})^2-b\frac{c}{b}+c)=0$

k can't be zero or y would be zero and this whole process would be moot. $e^{-\frac{c}{b}x}$ also can't be zero, so


Now, the middle term in the quadratic could easily collapse the equation. I could make the mistake here to simplify. Instead, I will avoid simplifying it and instead solve the quadratic with respect to $\frac{c}{b}$. To make this clearer, I will set $m = \frac{c}{b}$ so that I have



Now, let's say that this quadratic has two roots, $\alpha$ and $\beta$. Then



$\alpha+\beta = -\frac{b}{a}$


Substitute these two terms into the original equation gives me



expand this

$\frac{d^2y}{dx^2}-\frac{dy}{dx}\alpha - \frac{dy}{dx}\beta+\alpha\beta y=0$

rearrange and regroup

$\frac{d}{dx}(\frac{dy}{dx}-\beta y)= \alpha(\frac{dy}{dx}-\beta y)$

let's say that

$\frac{dy}{dx}-\beta y = z$


$\frac{dz}{dx}=\alpha z$

seperate the variables and integrate

$\int\frac{dz}{z}=\int\alpha dx$

$ln z = \alpha x + J$

where J is my unknown constant.

$Je^{\alpha x} = z = \frac{dy}{dx}-\beta y$

Solve this first order equation

$(e^{-\beta x}y)'=Je^{-\beta x}e^{\alpha x}$

$(e^{-\beta x}y)'=Je^{(\alpha-\beta) x}$

$e^{-\beta x}y = J\int e^{(\alpha-\beta) x} dx$

$e^{-\beta x}y = \frac{J}{\alpha-\beta}e^{(\alpha-\beta) x} + B$

where B is yet another unknown constant. Solving for y, I get

$y=\frac{J}{\alpha-\beta}e^{\beta x}e^{(\alpha-\beta)x}+Be^{\beta x}$

$y=\frac{J}{\alpha-\beta}e^{\alpha x}+Be^{\beta x}$

If I set $A=\frac{J}{\alpha - \beta}$, then

$y=Ae^{\alpha x}+Be^{\beta x}$

Which is the general solution, assuming the quadratic has two roots.

Hopefully, there are no typos.

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Well, it works. It seems like the most awkward workaround though to solve the ODE. I'm not sure how it would extend to higher dimensions (if it is meant to). – Daryl Sep 21 '12 at 4:20
A really complicated way to do something simple. Somewhere in the middle you conveniently forget that $m = c/b$ and you end up solving the quadratic $a m^2 - b m + c$ with $m$ as a variable, which you may as well have done in the first place. – Robert Israel Sep 21 '12 at 7:32
1) How complex you think it is has no bearing on the question, so that point is moot. 2) 'conveniently forgetting' was the entire point and, in fact, the subject of my original question. 3) You couldn't answer the question. A complicated solution is better than no solution. – Korgan Rivera Sep 21 '12 at 14:42
@Daryl The thing is that, even though it seems to work, my question still stands. If I treat m like a variable in the quadratic it works out. But I know that it's a constant, and the constant is not the same as the roots in a second order equation (although it works out with a first order equation.) So, I don't know why it works but it does. And that's just as frustrating. – Korgan Rivera Sep 23 '12 at 14:57
It almost looks like the approach you would take if $a,b,c$ are not constants, but simple functions of $x$. Something like a second order integrating factor, I'm not sure. – Daryl Sep 23 '12 at 19:57

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