Question about solving 2nd order linear differential equations

Question

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:

$a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=0$

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

$b\frac{dy}{dx}+cy=0$

Then I rearrange:

$\frac{dy}{dx}+\frac{c}{b}y=0$

The general solution of this is:

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

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:

$m=-\frac{c}{b}$

So the derivatives above become

$y=ke^{mx}$

$y' = kme^{mx}$

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

I then substitute this into my original equation to get:

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

or

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

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

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

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

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

$(am^2)=0$

or

$(\frac{ac^2}{b^2})=0$

$\frac{ac}{b}=0$

$ac=0$

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.

Answer 1

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 $a \ne 0$.

Answer 2

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 $a=0$.

Answer 3

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:

$a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=0$

set $a = 0$ (TEMPORARILY!)

$b\frac{dy}{dx}+cy=0$

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

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

simplify

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

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

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

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

$am^2-bm+c=0$

$m^2-\frac{b}{a}m+\frac{c}{a}=0$

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

$(m+\alpha)(m+\beta)=0$

$m^2+m(\alpha+\beta)+\alpha\beta=0$

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

$\alpha\beta=\frac{c}{a}$

Substitute these two terms into the original equation gives me

$\frac{d^2y}{dx^2}+\frac{b}{a}\frac{dy}{dx}+\frac{c}{a}y=0$

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

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

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

Then

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

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.