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

has derivative

, second derivative

, 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

.

If

, then

,

and, in general,

. Putting those into

**any** linear differential equation with constant coefficients will give you a polynomial in r times

. If that is equal to 0, since

is never 0, we can divide both sides by it, leaving that polynomial equal to 0.

For example, with

,

. And, as I said, since

is not 0, we can divide by it leaving

which has r= 1 and r= 2 as roots. That is, both

and

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...