Where does the general solution for second order linear ODEs come from?

[sandy.bridge]

If
$$ay+b\int^y_0ydy+cy'=0$$
then
$$ay'+by+cy''=0$$
now, let
$$y=e^{sx}$$
thus,
$$s^2+a/cs+b/c=0$$
and then one solves for s. It is then plugged into what sources are deeming a "general solution"
$$y=C_1e^{s_1x}+C_2e^{s_2x}$$
however, none of these texbooks explain or derive where this comes from, and I have not taken any linear algebra or differential equations. Can someone explain where $$y=C_1e^{s_1x}+C_2e^{s_2x}$$ comes from?
Thanks!

 

[chiro]

If
$$ay+b\int^y_0ydy+cy'=0$$
then
$$ay'+by+cy''=0$$
now, let
$$y=e^{sx}$$
thus,
$$s^2+a/cs+b/c=0$$
and then one solves for s. It is then plugged into what sources are deeming a "general solution"
$$y=C_1e^{s_1x}+C_2e^{s_2x}$$
however, none of these texbooks explain or derive where this comes from, and I have not taken any linear algebra or differential equations. Can someone explain where $$y=C_1e^{s_1x}+C_2e^{s_2x}$$ comes from?
Thanks!

Hey sandy.bridge.

Lets take it from the step if y = e^mx, y' = me^mx, y'' = m^2 e^mx then you end up getting an equation of the form:

(am^2 + bm + c) x e^(mx) = 0. Now divide by e^(mx) which gives you the quadratic term:

am^2 + bm + c = 0.

Now if you have two solutions for m, what does that imply for the final solution?

The real 'guts' of proving this general property for linear ODE's or systems of ODE's is based on the superposition principle. The idea uses linear algebra, in particular spectral theory to show the solution in terms of superpositions. If you want to prove it yourself, you need to setup the ODE in terms of a matrix system, and then do some eigen-analysis to decompose the matrix.

I didn't have to do any proving when I took my prior DE course, but if you want to understand why you get this kind of formula in a general sense, look at the superposition principle with reference to ODE's and a detailed proof of it.

Also I didn't link to Wikipedia since they are having a blackout for SOPA awareness.

 

[HallsofIvy]

Note that while "trying" $e^{rx}$ will reduce a linear differential equation with constant coefficients to its "characteristic equation", a polynomial, that does not necessarily mean the solution will involve exponentials.

If the solution to the characteristic equation is a complex number, a+ bi, then (assuming all coefficients are real) a- bi is also a root and $e^{(a+ bi)x}= e^{ax}\left(cos(bx)+ sin(bx)\right)$. Also, if r is a multiple root (If the differential equation is y'''+ 3y''+ 3y'+ y= 0, the characteristic equation is $r^3+ 3r^2+ 3r+ 1= (r+1)^3= 0$ which has r= -1 as a triple root. The general solution to the differential equation is $$y= Ae^{-x}+ Bxe^{-x}+ Dx^2e^{-x}$$.