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Homework Help: Seperable Differential Equation

  1. Jul 22, 2009 #1
    1. The problem statement, all variables and given/known data

    I have dP/dt = (b - kP)P - hP (population problem with harvesting)

    I'm told this is separable into a 1st order DE.... so this is what I've tried:

    dP/dt = bP - kP^2 - hP
    dP/dt + (h-b)P = -kP^2
    At this point, I get stuck... It doesn't appear to be a separable DE to me... I've tried approaching this as a Bernoulli equation...

    let Z = P^-1
    dZ/dt = (-1)(p^-2)(dP/dt)
    (-P^2)(dZ/dt) = dP/dt

    therefore, substituting back into the 3rd equation:

    (-P^2)(dZ/dt) + (h-b)P^-1 = -kP^-2
    dZ/dt + (h-b)P = -k

    let mu = e^[Integral (h-b)P^-1 dp] = e^[(h-b) ln P] = P^(h-b) (we don't care about the "+c" since it falls out in the next step...)

    Multiplying the 1st order DE by mu
    P^(h-b)*dZ/dt + [P^(h-b)](h-b)P^-1 = -kP^(h-b)

    d/dt [ P^(h-b) * Z] = -kP^(h-b). Integrating both sides gives:
    P^(h-b) * Z = -ktP^(h-b). Substituting Z = P^-1 back in:
    P^(h-b) * P^-1 = -ktP^(h-b) + C
    P^-1 = -kt + C / P^(h-b)
    P = 1 / [-kt + C/P^(h-b)]

    Is this right? It seems horribly wrong, given this is a logistic equation... Any help in spotting my mistake would be greatly appreciated. I've been told you can just divide both sides of the original equation to get dP/[kP^2 +(h-b)P = dt. However, in order to integrate that, would you need to complete the square on the bottom?
    Last edited: Jul 22, 2009
  2. jcsd
  3. Jul 22, 2009 #2


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    Homework Helper

    I have not read through your entire solution, but you can separate it as follows (assuming b,k,h are constants).

    \frac{dP}{dt}= (b - kP)P - hP \Rightarrow \frac{dP}{(b - kP)P - hP}=dt

    Now integrate both sides.
    Last edited: Jul 22, 2009
  4. Jul 22, 2009 #3


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    It's already first order!

    What do you understand separable to mean?

    Your attempted answer is I'm afraid really going about it the wrong way! But I wanted to point out some actual mistakes in it anyway...

    Some of those P's should be Z's in the last expression quoted above. This error carries through your working...

    Another error. The second line is not equivalent to the first line (if you carry out the d/dt correctly you'll see it has P derivatives in it).

    You really need to be careful setting things up like this. Normally the point of a substitution is to eliminate one variable for an (easier to deal with) alternative. However your substitution of Z for P has left you with an equation with both P's and Z's in. This should be telling you it wasn't a good substitution!

    That's a much better way to go about it, and is what is normally meant by "separable" in this context.

    Not necessarily --- can you think of any other technique you have learnt for dealing with fractions involving polynomials? Let me know if you need a hint...
  5. Jul 22, 2009 #4
    Oh wow... Mistakes everywhere! Thanks for the help, guys!

    jmb, I'm guessing, you're hinting at partial fractions, right? In that case, I'm assuming I'd have to change the bottom of the left hand side from:

    bP - hP - kP^2



    and then separate this by partial fractions for

    1/[(b-h-kP)(P)] = A(P)/[(b-h-kP)(P)] + B(b-h-kP)/[(b-h-kP)(P)]

    which implies

    1 = Bb - Bh
    0 = AP - BkP

    and thus, B = 1/(b-h), A = k/(b-h), which can then be used to integrate the left hand side.

    Seem good so far?

    Thanks again :)
  6. Jul 22, 2009 #5


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    Looks good. Let us know how you get on with the integration.
  7. Jul 22, 2009 #6
    Alright! Here goes:

    k/(b-h) Integral 1/(b-h-kP) dP + 1/(b-h) Integral 1/P dP = Integral 1 dt

    -1/(b-h) ln |b-h-kP| + 1/(b-h) ln |P| = t + c

    Pulling out the 1/(b-h) term, and combining the natural log terms:

    1/(b-h) ln|P/(b-h-kP)| = t + c

    Multiplying by (b-h) and raising both sides by e:

    P/(b-h-kP) = e^[t(b-h)+c(b-h)]
    P/(b-h-kP) = Ce^(t(b-h))

    Multiplying both sides by (b-h-kP) and collecting like terms:

    P + kPCe^(t(b-h)) = bCe^(t(b-h)) - hCe^(t(b-h))

    P (1 + kPCe^(t(b-h))) = (b-h) Ce^(t(b-h))

    dividing to get:

    P = (b-h) Ce^(t(b-h)) / (1 + kPCe^(t(b-h)))

    And finally, multiplying both terms by e^(-t(b-h))

    P = (b-h)C / [e^(-t(b-h)) + kC]

    Whew!! Look good so far? :) If so, I have one additional question that relates to this:

    We also have to analyze what happens to the population when h > b (harvesting > birth rate)... So, I began by setting

    dP/dt = 0 (equilibrium points) = (b - h - kP)P
    which gives the points P = 0, P = (h-b)/k

    Now, looking at an "arbitrary" Po, I check the cases:

    Po > (h-b) / k implies dP/dt < 0
    0 < Po < (h-b) / k implies dP/dt > 0

    This seems to say that P = (h-b)/k is a stable equilibrium point. (I know this isn't rigorous, but this is the method we're using in class...) However, given the nature of the problem, that can't be right... It's saying that if I harvest more of the population than gives birth, then I will still reach equilibrium. Where am I going wrong?
  8. Jul 23, 2009 #7


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    I haven't checked through your working line by line, but that agrees with my own answer. Note though that it is more conventional in time dependent problems to write the constant of integration [tex]C[/tex] in terms of the value of [tex]P[/tex] at [tex]t=0[/tex] (often written [tex]P_0[/tex]).

    You've made a sign error in calculating the equilibrium points. Correct this and then see if you can make more sense of things...
  9. Jul 23, 2009 #8
    Ok, I must be stupid or something, because I'm still horribly confused....

    Even taking into account that the critical points are actually P = 0, and P = (b-h)/k (sign error removed!), the solution still doesn't seem to make sense.

    Po > (b-h)/k implies dP/dt > 0
    Po < (b-h)/k implied dP/dt < 0

    But, if I'm harvesting more (fish in this example) than are being born, shouldn't they die out eventually, instead of having a population explosion (as in the first case)? Or is it because it's an unstable equilibrium that I don't have to worry about this case?
  10. Jul 23, 2009 #9
    Wait a tick! If P = (b-h)/k and h > b, that implies that P is negative. Since we cannot have a negative population, we cannot consider this case... Therefore, we only care about the case where Po > 0, in which case dP/dt < 0.

    How does that sound?? :)

    Thanks again for all of the help!
    Last edited: Jul 23, 2009
  11. Jul 24, 2009 #10


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    Yes. If [tex]h>b[/tex] then, for all physical values of [tex]P[/tex] (i.e. [tex]P>0[/tex]) you have [tex]dP/dt < 0[/tex], and so the population will evolve towards the equilibrium point at [tex]P=0[/tex].

    As a bonus question, can you tell me from your solution at what value of [tex]t[/tex] the system will reach this equilibrium point ([tex]P=0[/tex])?

    Also, in the interests of completeness, there is a choice of [tex]h,b[/tex] that invalidates your solution: what is this choice, and what would the solution be in that case?
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