# Homogeneous second order linear ODE

Gold Member
Hey; not much of a homework question, but something i was wanting to find out.

Im still a first year undergraduate and just started on differential equations. We have just finished going over homogeneous 2nd order ODE's of the form:

ay'' + by' + cy = 0

My texbook briefly outlines the solution to any general equation, and from there on simply uses the derived solutions to solve further problems.

The way in which it derives the general solution is by noting that if we let y=e^rx, then the equation becomes:

$$ar^2 e^{rx} + bre^{rx} + ce^{rx} = e^{rx} (ar^2 + br + c) = 0$$

And since exp(x) is real and positive for all x, the equation only holds true if:

$$ar^2 + br + c = 0$$

Now that all makes sense to me, but what isnt really clear to me is why we just happen to select y=e^rx? Why not some other function? Is there some proof that shows e^rx as the only function to satisfy the conditions? I realize that e^rx is ideal because of the special property that its derivatives are all scalar multiples of itself, but what reason is there to deny the existence of some other function with similar properties?

Thanks for the help; Even a link pointing me in the right direction is apreciated

Dan.

The idea is that you need some function whose derivatives are not linearly independent, that is, dy/dx = cy. This is separable, that is, dy/y = cdx. Solving yields ln |y| = cx + d. Now let us rearrange to get |y| = (e^d)e^(cx). We can eliminate the absolute value by allowing the e^d term to be negative... call it k. Then

y = k e^(cx).

That's where it comes from, and it is the only function which has derivatives which are not linearly independent to each other and the original function (except one other class of functions: which?)

HallsofIvy
Roughly speaking, In order that $ad^2y/dx^2+ b dy/dx+ cy= 0$, the different derivatives must cancel and, since the coefficients are constants, that means the derivatives must be of the same type as the function y itself. For example, if y= ln(x), then dy/dx= 1/x and $d^2y/dx^2= -1/x^2$ so there is nothing to cancel the "ln x" in the cy term. Of course, exponentials, ekx is the simplest function whose derivative, kekx, is the same type of function.