# How do you solve second-order linear differential equations without inspection?

In most explanations, you are given something like

$ay''+by'+cy=0$

and the method starts with something like, "I can see that the exponential function would be a solution, and also adding two exponential functions together would work, so..." and so on.

What if I can't just magically see that this solution covers every possible solution? How do you find a solution for this kind of equation algebraically? I thought I had come up with a solution in an earlier question (http://math.stackexchange.com/a/200082/21050), but I didn't understand why my solution worked so that's no good.

I just learned a clever way to solve these when $c=0$, but assume that $c$ is not zero for your answer, please.

edit: I want to alter this question slightly: I can go ahead and assume that $e^{mx}$ is a solution and work from there. But what if I was braindead and couldn't make that guess? How can I get $e^{mx}$ to spit out somewhere?

-

You observe that the derivative of $\left(\begin{matrix}y\\y'\end{matrix}\right)$ is $\left(\begin{matrix}y'\\y''\end{matrix}\right)=\left(\begin{matrix}y'\\\frac{-cy-by'}{a}\end{matrix}\right)$ (assuming $a\ne0$), that is differentiation looks like multiplication with $A=\left(\begin{matrix}0&1\\-\frac ba&-\frac ca\end{matrix}\right)$. With a suitable basis change this becomes diagonal (except in one special case) and hence we are led to search for eigenfunctions under differentiation - but that is sometimes the very definition of $e^x$. It is instructive to find out, what the exceptional case I mentioned loooks like and how we are led from the matrix to use a function looking like $x e^x$ as basis element.

-
I followed you as far as eigenfunctions and now I have to revise linear algebra. –  Korgan Rivera Sep 28 '12 at 18:29

I'm not sure how well this works with roots that aren't real, but as long as a,b, and c are constant, it's possible to treat the left side similar to a polynomial. Let's say that

$$x^2+\frac{bx}a+\frac ca=(x-m)(x-n)$$

Then it is possible to rewrite $y''+\frac{by'}a+\frac{cy}a$ as $(y'+my)'+n(y'+my)$ or $(y'+ny)'+m(y'+ny)$. From here, it can be solved through subtstitution and integrating factors.

Let's take an example of $y''+10y'+25y=0$. $x^2+10x+25=(x+5)^2$ and our equation can be rewritten as

$$y''+5y'+5y'+25y=(y+5y)'+5(y'+5y)=0$$

Here we make a substitution

$$u=y'+5y$$

$$u'+5u=0$$

I'll assume you know how to solve this type of equation via integrating factors.

$$e^{5x}u'+5e^{5x}u=(e^{5x}u)'=0$$

$$e^{5x}u=k_1$$

$$u=k_1e^{-5x}$$

$$y'+5y=k_1e^{-5x}$$

The same integrating factor works here, which will lead to $e^{5x}y=k_1x+k_2$. Double roots are not a special case with this method and the particular solution is calculated directly.

So, to sum up, you can multiply the equation through by an integrating factor of the form $e^{mx}$; integrate both sides, introducing a constant of integration; then divide back through by that integrating factor.

-