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**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...