# Second order ODE's, proof of solution method

• Dec 11th 2011, 05:31 AM
TwoPlusTwo
Second order ODE's, proof of solution method
When solving second order ODE's, we always assume that all solutions will be proportional to e^rt.

How hard is it to prove that this has to be the case?

If there's a simple proof, please don't just state it. In stead give me a hint of where to begin, so I can try to do it myself.

If the proof is long and complicated, I don't need to know all the details (because I wouldn't understand them anyway). But it still would be nice to get a brief outline of what it involves.
• Dec 11th 2011, 06:30 AM
Jester
Re: Second order ODE's, proof of solution method
Here's one way but I don't know if it's a proof or not. Consider the ODE

$\displaystyle y'' -(a+b)y' + aby = 0,\;\;\; a\ne b \;\;(1)$

where $\displaystyle a$ and $\displaystyle b$ are constant. I'm assuming that you meant constant coefficients otherwise the solutions are not usually in the form $\displaystyle e^{rt}$. Now we will factor this equation using operators. We can re-write $\displaystyle (1)$ as

$\displaystyle \left(\dfrac{d}{dt} - a\right)\left(\dfrac{d}{dt} - b\right) y = 0. \;\; (2)$.

If we let

$\displaystyle u = \left(\dfrac{d}{dt} - b\right) y$
then (2) becomes

$\displaystyle \left(\dfrac{d}{dt} - a\right)u = 0, (2)$.

or

$\displaystyle u' - au = 0$.

We solve (it's separable) giving

$\displaystyle u = c_1 e^{at}$

Now

$\displaystyle \left(\dfrac{d}{dt} - b\right) y = u = c_1 e^{at}$

or

$\displaystyle y' - by = c_1 e^{at}$

This is linear which we integrate giving

$\displaystyle y = c_1 e^{at} + c_2 e^{bt}$

noting that I have absorbed some constants into $\displaystyle c_1$.