Diff Equation with two populations

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The discussion focuses on solving a differential equation involving two populations, specifically addressing the equation dP1/dt = kP1 + M1. Participants explore the integration of this linear equation, leading to an exponential function of time. The importance of determining the constant k is highlighted, with suggestions to use initial conditions to find specific solutions for different time intervals. Clarifications are made regarding the use of arbitrary constants in the integration process and how to incorporate them into the general solution. Ultimately, the conversation emphasizes the need to combine solutions from the homogeneous and inhomogeneous equations to achieve the final result.
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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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