# Differential equations (application)

1. Apr 24, 2014

### delsoo

1. The problem statement, all variables and given/known data

hi, i have difficulties in this question... can you teach me how to get the ans please... i dont have the ans . this involved differential equations

2. Relevant equations

3. The attempt at a solution

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2. Apr 24, 2014

### D.W.

So... Quick note for the beginning step of the problem. You've defined x as the number of rabbits at a given time. But, remember, x is a function of t - x changes with time. You correctly set up the differential equation (i.e. dx(t)/dt=kx(t) ), but just stare at that real quick and see what it tells you. You have a function of t whose derivative with respect to t is just the same function multiplied by k. What function has the property that when you differentiate it, you get back the same thing, multiplied by a constant?

3. Apr 24, 2014

### delsoo

How should o proceed then??

4. Apr 24, 2014

### HallsofIvy

First you are told that the reproduction rate of the rabbits is proportional to the number of rabbits. Yes, that is the same as dx/dt= kx. You can use the fact that the number of rabbits doubled in 5 years (60 months) to determine k.

But then you are told that "an outbreak of a certain disease caused the death of 100 rabbits per month". I see that you have let "y" be the "number of deaths". Since that is constant at 100 per month, I wouldn't do that. Rather, I would say, as long as t is the time measured in months, dx/dt= kx- 100. And, since we are told, at the time this disease began the rabbits had "doubled to 10000", I would take t= 0 at the time the disease began and x(0)= 10000, not 5000.

Solve that equation for x(t). Then, since you are asked for the number of rabbits two years after the outbreak of the disease, and t is in months, find x(24).

Last edited by a moderator: Apr 24, 2014
5. Apr 24, 2014

### delsoo

do u mean this?

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6. Apr 26, 2014

### delsoo

do u mean this? i have redo the question...

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7. Apr 26, 2014

### HallsofIvy

No, I don't mean either of those. In the last you have the equation dx/dt= kx when I told you that you that the equation should be dx/dt= kx- 100 (with t measured in months). In the first, you start with the equation dx/dt= kx- 100 but then have "dx/dt= (1/5) ln 2(10000)- 100(24)". I don't know where you got that!

Do you know how to solve an equation of the form dx/dt= kx- 100?

8. Apr 28, 2014

### delsoo

sorry i may i know how to solve it please? i'd been thinking of this quite long . this is driving me crazy!!!