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Complex differential equations to find functions [TRIED]

  1. Apr 24, 2013 #1
    1. The problem statement, all variables and given/known data

    I need to solve two differential equations to find a population function P(t). I am able to do this with problems like newtons law of cooling:

    dT/dt=-k(T-Ta) solves to:
    dT=-k(T-Ta) dt
    ∫1/(T-Ta)dT=∫-k dt
    Ln(T-Ta)=-kt
    e-kt=T-Ta
    T=Ta+E-kt

    However I have been presented with two hard problems which I can not do:

    1. dP/dt=kP-AP2
    2. dP/dt=kP(M-P)

    using P=Po at t=0 find P(t) for both.

    2. Relevant equations

    ^As above, no other equations where provided, it is pretty much calculus and algebra

    3. The attempt at a solution

    dP=kP-AP2 dt
    dP=P(k-AP) dt
    1/P dP = k-AP dt
    ∫1/P dP = ∫k-AP dt
    Ln(P)=t(k-AP)
    Ln(P)=tk-APt
    etk-APT=P
    loge(tk-APt)=P
    eP=tk-APt

    From there it just loops around. I can not think of anything else to do with it. As you can not (that I am aware of, unless you used unreals, factorise kP-AP2.

    Thanks for reading this, I am not giving up and will continue to work on it now.
     
  2. jcsd
  3. Apr 24, 2013 #2

    LCKurtz

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    $$
    \frac{dP}{P(k-AP)} = \frac {-1} A \frac {dP}{P(P-\frac k A)}$$Use partial fractions.
     
  4. Apr 25, 2013 #3
    Thanks a lot, I have now come down to: XP2- XPk/A - AY=1 (Instead of using A and B for partial fractions I used X and Y as those variables were not already used). The questions we have done in class involve something like: x-3=Ax-A+Bx+B and we just split them up into terms which contain x and terms not containing X to solve for A and B. Would you split it up into terms containing P and not P or terms not containing P or k or A?

    We have not done this type of question containing two unknown variables (apart from those used for the partial fractions of course) yet. I apologize that I do now know how to use MathsJax, I am just using the quick symbols.

    EDIT: Should I have used three fractions and used X,Y,Z as P(k-AP) could be factorized to:

    -A times P times P - k/a. Then split the the final line into variables containing each one (somehow, some might contain two variables but I suppose they could be eliminated when finding the values for X, Y and Z). I will try it now...
     
    Last edited: Apr 25, 2013
  5. Apr 25, 2013 #4

    LCKurtz

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    Just call the P's x's and call -k/A "c" and you will feel right at home$$
    \frac 1 {x(x-c)}$$with a -1/A out in front.
     
  6. Apr 25, 2013 #5
    That makes a lot more sense, I am doing it now. I assume the -1/A can be simply integrated to -Ln(A) as you just simplified -1/A to -1 times 1/A. I will post my results once I finish, thank you a lot by the way. I would be unable to do this without your help.
     
  7. Apr 25, 2013 #6
    This is my working:

    ∫1/(Pk- AP2) dP
    = ∫-1/A times p(p-k/a) dP
    ∫-1/A dP can be solved
    = -1P/A (this could be incorrect)
    now back to the other part (I know this setting out is terrible)
    ∫ p(p-k/a) dP

    let P=x and -k/A=c

    ∫ x(x-c) dx
    ∫ A/x +B/(x-c) dx
    ∫ A(x-c)/x(x-c) +Bx/x(x-c) dx
    ∫ A(x-c)+Bx / x(x-c) dx
    ∫ Ax-Ac+Bx / x(x-c) dx

    Therefore Ax-Ac+Bx / x(x-c) = 1
    Split up into terms containing x and terms not containing x

    X
    A+B=0
    B=-A

    Other
    -Ac=1
    -A=1/c
    A=-1/c

    Therefore B= 1/c

    -1/c / x + 1/c / (x-c) = 1 / Pk-AP2
    Sub C and P back in...
    A=-1/c
    A=-1/(-k/A)
    A =A/k

    B=1/c
    B=1/(-k/A)
    B=-A/k

    So you get:

    ∫-1/A times [A/k / P + -A/k / (P + k/A) ] dP
     
  8. Apr 25, 2013 #7

    LCKurtz

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    Why would you assume that? -1/ A is just a constant out in front. The variable of integration is P (or x if you renamed it).
     
  9. Apr 25, 2013 #8
    Oh, that makes sense. My bad, it was lake when I typed that. Is my working for the A and B values correct, I will try to differentiate the final line now.
     
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