Differential equations (application)

[delsoo]

Homework Statement

hi, i have difficulties in this question... can you teach me how to get the ans please... i don't have the ans . this involved differential equations

Homework Equations

The Attempt at a Solution

 

[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?

 

[delsoo]

How should o proceed then??

 

[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).

 

[delsoo]

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).

do u mean this?

 

[delsoo]

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).

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

 

[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?

 

[delsoo]

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