Diff Equation with two populations

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Homework Help Overview

The discussion revolves around a differential equation involving two populations, specifically focusing on the dynamics described by the equation dP1/dt = kP1 + M1. Participants are exploring the implications of the constant k and the initial conditions for solving the equation.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss integrating the linear equation and the implications of the constant k. There are attempts to clarify the relationship between constants derived from integration and those introduced in the problem. Some participants suggest using separation of variables and initial conditions to find specific solutions.

Discussion Status

Several participants are actively engaging with the problem, offering different methods for approaching the solution. There is a mix of interpretations regarding the constants involved and how to apply initial conditions. Guidance has been provided on integrating the equation and considering both homogeneous and inhomogeneous solutions.

Contextual Notes

There are mentions of specific initial conditions, such as P(0) = 1000, and the need to evaluate solutions over different intervals of time. Participants are also questioning the assumptions related to the constants and their roles in the equations.

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Homework Statement


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Homework Equations


Not sure


The Attempt at a Solution


I tried to do the first population problem at t=0 so M(t) would be 100, but there is the k constant that I don't know and don't know how to find it. Is there a way to find k or do you even need too?
 
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You have dP1/dt= kP1+ M1. That is a linear equation so we can divide it into two parts. dP1/dt= kP1 can be itegrated by rewriting it as dP1/P1= k dt and integrating both sides. You should find that P is an exponential function of t.

Then consider dP1/dt= kP1- 100 for t between 0 and 1. Looking for a constant soluton, P1= C, the derivative is 0 so we must have 0= kC- 100 of C= 100/k. Adding that to the exponential gives the general solution for 0< t< 1. To find the solution for t between 1 and 2, evaluate the first solution at t= 1 to get a condition so you can solve the same equation using that condition.

Yes, your solutions will be functions of k, not specific numbers.
 
I got the exponential equation to be e^(kt)e^c

Now you said to add the C to the exponential. I'm assuming the C your talking about is not the same as the arbitrary constant that I get from integration. So do you want me to literally add it on or what?
 
Colts said:
I got the exponential equation to be e^(kt)e^c

Now you said to add the C to the exponential. I'm assuming the C your talking about is not the same as the arbitrary constant that I get from integration. So do you want me to literally add it on or what?

If you don't follow Hall's argument just take the straightforward approach and solve the differential equation on 0<=t<=1 using separation of variables. Use the initial condition P(0)=1000 to find the constant and figure out what P(1) is. Use that for an initial condition on 1<=t<=2.
 
Halls showed how to solve dP/dt= kP but you need to solve dP/dt = kP + M. Since the equation is linear, you can get the general solution you want by adding the solution you found to the homogeneous equation to any solution of the complete, inhomogeneous one. In this case, it's easy to see that one solution is P constant. Halls calls this constant C. So you write dP/dt = 0 and get C = -M/k. (Halls had M1 as an emigration rate of 100, so had the wrong sign.)
Putting this together, P = Aekt - M/k. Now you can plug in the initial conditions to find A.
 

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