# Thread: Determining what values of r have solutions of the form y= e^(rt)

1. ## Determining what values of r have solutions of the form y= e^(rt)

It's been a while since I've touched calculus and I have these two problems I'm having trouble starting on. Can anyone help me understand what steps I need to take to solve these two problems please:

Determine the values of r for which the differential equation y'' - 3y' + 2y = 0 has solutions of the form y = e^(rt). I saw a similar problem earlier and they differentiated y = e^(rt) to get y' = r * e^(rt) but I don't understand why that step was taken.

A population satisfies the differential equation dp/dt = 0.5p - 450, where t is measure in months. Find the time at which the population becomes extinct if p(0) = 850.
Not sure where to start here...

2. ## Re: Determining what values of r have solutions of the form y= e^(rt)

Originally Posted by colerelm1
It's been a while since I've touched calculus and I have these two problems I'm having trouble starting on. Can anyone help me understand what steps I need to take to solve these two problems please:

Determine the values of r for which the differential equation y'' - 3y' + 2y = 0 has solutions of the form y = e^(rt). I saw a similar problem earlier and they differentiated y = e^(rt) to get y' = r * e^(rt) but I don't understand why that step was taken.
The function $\displaystyle e^{rx}$ has derivative $\displaystyle re^{rx}$, second derivative $\displaystyle re^{rx}$, etc. That is, every derivative is a multiple of the original function. On other hand an equation like y''- 3y'+ 2y= 0 has only numerical multiples of the derivatives adding to 0. In order that they cancel, the various derivatives must be the same "type" of function. That should suggest that when we have an equation that involves derivatives, multiplied only by constants, we look for solutions of the form $\displaystyle e^{rx}$.

If $\displaystyle y= e^{rx}$, then $\displaystyle y'= re^{rx}$, $\displaystyle y''= r^2e^{rx}$ and, in general, $\displaystyle y^{(n)}= r^ne^{rx}$. Putting those into any linear differential equation with constant coefficients will give you a polynomial in r times $\displaystyle e^{rx}$. If that is equal to 0, since $\displaystyle e^{rx}$ is never 0, we can divide both sides by it, leaving that polynomial equal to 0.

For example, with $\displaystyle y= e^{rx}$, $\displaystyle y''- 3y'+ 2y= r^2e^{rx}- 3(re^{rx})+ 2e^{rx})= (r^2- 3r+ 2)e^{rx}= 0$. And, as I said, since $\displaystyle e^{rx}$ is not 0, we can divide by it leaving $\displaystyle r^2- 3r+ 2= (r- 2)(r- 1)= 0$ which has r= 1 and r= 2 as roots. That is, both $\displaystyle y= e^{x}$ and $\displaystyle y= e^{2x}$ satisfy the differential equation.

A population satisfies the differential equation dp/dt = 0.5p - 450, where t is measure in months. Find the time at which the population becomes extinct if p(0) = 850.
Not sure where to start here...
That's a straight forward "separable" first order differential equation. You should have learned to solve such an equation long before "higher order linear equations with constant coefficients" as in the first problem. You can write it as $\displaystyle \frac{dp}{0.5p- 450}= dt$ and integrate both sides. What do you get when you integrate?

3. ## Re: Determining what values of r have solutions of the form y= e^(rt)

Ok, thanks. I worked it out and I got .5*ln|p-900|. How do I figure out at what time the population becomes extinct?