# Rodent Population as a differential equation.

1. Jan 12, 2014

### cp255

1. The problem statement, all variables and given/known data
https://www.physicsforums.com/attachment.php?attachmentid=65556&stc=1&d=1389570667

For those who can not see the screen shot here is the question...
Suppose the population P of rodents satisfies the diff eq dP/dt = kP^2.
Initially there are P(0) = 2 rodents, and their number is increasing at the rate of dP/dt = 1 rodent per month when P = 10. How long does it take for the population to reach 105 rodents.

2. Relevant equations

3. The attempt at a solution

First I found k by substituting in 10 for p and 1 for dp/dt.
1 = k * 10^2
k = 1/100

Then to solve the differential equation I integrated both sides with respect to t.

∫dp/dt * dt = ∫0.01 * p^2 dt
p = 0.01p^2 * t + C

I solved for C and found C = 2.

Solving for t gives
t = 100(p - 2) / p^2

I then plunged in 105 for p and the answer was wrong.

Last edited: Jan 12, 2014
2. Jan 12, 2014

### 1MileCrash

It would be helpful to see the question.

3. Jan 12, 2014

### epenguin

At the moment the question has not been posted or is not visible to me.

4. Jan 12, 2014

### cp255

I can see the screen shot of the question uploaded. I will rewrite the question and update the original post.

5. Jan 12, 2014

### Dick

Integrating ∫0.01 * p^2 dt to get 0.01p^2*t+C is incorrect. That's assuming p is a constant. You need to separate the ODE first.

6. Jan 13, 2014

### HallsofIvy

Staff Emeritus
τ
This is incorrect. You are treating p, on the right, as if it were a constant but it is a function of t.
Instead, write it as $$\int \frac{dp}{p^2}dp= \int .01dt$$
$$-\frac{1}{p}= .01t+ C$$
$$p(t)= -\frac{1}{.01t+ C}$$
$$p(0)= -\frac{1}{C}= 2$$