- #1

meggs521

- 9

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So here's the problem:

**"Suppose the number x(t) (with t in months) of alligators in a swamp satisfies the differential equation**

dp/dt = 0.0001x^2 - 0.01x

(a) If there are initially 25 alligators in the swamp, solve this differential equation to determine what happens to the alligator population in the long run.

(b) Repeat for an initial population of 150 alligators."

dp/dt = 0.0001x^2 - 0.01x

(a) If there are initially 25 alligators in the swamp, solve this differential equation to determine what happens to the alligator population in the long run.

(b) Repeat for an initial population of 150 alligators."

The main part that's throwing me is the dp/dt = stuff with x, but they say that x is the number of alligators, which basically means population, right? So should I be setting this up as dx/dt = 0.0001x^2 - 0.01x instead?

I did that and I kept getting the ln of a negative number when x(0) was 25 [My actual equation was: 100(ln(x-100)-ln(x)) = t+C]. And lns of negative numbers don't work, so I assumed it's not supposed to be dx/dt = 0.0001x^2 - 0.01x.

But when I use dp/dt = 0.0001x^2 - 0.01x and integrate I get:

P = (0.0001x^2 - 0.01x)t + C

If this is the right way to do it then I know that if t is zero then x doesn't matter and I can solve for C, which gives me C = 25, but this doesn't seem right. If this is right then how do I find the population in the long run? I need to find when the population dies out, so P would be 0 and I would solve for t to find the time. But with my second version I still have x that I have no values for.

I know I'm making some really obvious mistake and I just can't see it yet, but I'm completely lost and any help would be super! All of the examples I've found online or in the book are dp/dt= stuff with p. I remember doing problems like the one in the HW problem, but I can't remember how I did them.

Thanks!