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Math Help - Getting the general solution by using partial fractions [Differential Equation]

  1. #1
    DVS
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    Getting the general solution by using partial fractions [Differential Equation]

    Hi all,

    I'm not really confident with partial fractions, so this question really has me stumped at the moment.

    For (factorized form of Constant Harvest/Constant Effort models):

    \frac{dP}{dt} = \frac{r}{K} (P-Pu)(Ps-P)

    How would I use variable separation and partial fractions to find the general solution?

    So far I've made it to:

    \frac{dP}{(P-Pu)(Ps-P)} = \frac{r}{K} dt


    Am I on the right track?

    P does not have to be the subject.

    Thank you!
    Last edited by DVS; September 30th 2012 at 11:43 PM.
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    Re: Getting the general solution by using partial fractions [Differential Equation]

    Quote Originally Posted by DVS View Post
    Hi all,

    I'm not really confident with partial fractions, so this question really has me stumped at the moment.

    For (factorized form of Constant Harvest/Constant Effort models):

    \frac{dP}{dt} = \frac{r}{K} (P-Pu)(Ps-P)

    How would I use variable separation and partial fractions to find the general solution?

    So far I've made it to:

    \frac{dP}{(P-Pu)(Ps-P)} = \frac{r}{K} dt


    Am I on the right track?

    P does not have to be the subject.

    Thank you!
    Are you using \displaystyle \begin{align*} Pu \end{align*} to represent \displaystyle \begin{align*} P\cdot u \end{align*} or \displaystyle \begin{align*} P_u \end{align*}? Same for Ps?
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    Re: Getting the general solution by using partial fractions [Differential Equation]

    Hey DVS.

    The first thing I recommend is to take P outside of both terms. Doing this will give you:

    [1/P^2]*[1-u)^(-1)*[s-1]^(-1) which means you don't have to do a thing since s and u are independent variables of the one that is being integrating (i.e. you treat this as [1/P^2]*d where d is a constant.

    So the integral of d*[1/P^2] is just -d/P + C for an indefinite integral but you can carry this to the RHS and put the C there.
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    DVS
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    Re: Getting the general solution by using partial fractions [Differential Equation]

    Quote Originally Posted by Prove It View Post
    Are you using \displaystyle \begin{align*} Pu \end{align*} to represent \displaystyle \begin{align*} P\cdot u \end{align*} or \displaystyle \begin{align*} P_u \end{align*}? Same for Ps?
    Apologies for not mentioning that.

    P_{u} is the unstable equilibrium
    P_{s} is the stable equilibrium

    So it would be P_{u} , not P\cdot u
    Last edited by DVS; October 1st 2012 at 12:16 AM.
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    Re: Getting the general solution by using partial fractions [Differential Equation]

    You just have to rewrite 1/((P-Pu)(Ps-P)) on the form = (A/(P-Pu))+(B/(P-Ps)) with A, B to be expressed as functions on Pu and Ps.
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    DVS
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    Re: Getting the general solution by using partial fractions [Differential Equation]

    Had another look at it earlier,

    I now have:

    \frac{1}{(P-P_{u})(P_{s}-P)} = \frac{A}{(P-P_{u})} + \frac{B}{(P_{s}-P)}

    Then as I believe, I am supposed to multiply both sides by the denominator of the left side, which gives me:

     1 = \frac{A}{(P-P_{u})} (P - P_{u})(P_{s} - P) + \frac {B}{(P_{s} - P)} (P-P_{u})(P_{s} - P)

    But I don't see what that gets me or where I go to find the general solution from here.
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    Re: Getting the general solution by using partial fractions [Differential Equation]

    First, simplify. Then expand.
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    DVS
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    Re: Getting the general solution by using partial fractions [Differential Equation]

    Now:

     1 = A(Ps-P) + B(P-Pu)

    But I still don't understand - how does expanding that help?
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    Re: Getting the general solution by using partial fractions [Differential Equation]

    1 = B*P - A*P +A*Ps -B*Pu
    What is the relationship between B and A so that the expression be constant for any value of P ?
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    Re: Getting the general solution by using partial fractions [Differential Equation]

    Quote Originally Posted by DVS View Post
    Now:

     1 = A(Ps-P) + B(P-Pu)

    But I still don't understand - how does expanding that help?
    You want the two polynomials 1 and A(P_s-P) + B(P-P_u) to be equal. That means that all the coefficients of P have to be equal. To figure that out, you need to expand A(P_s-P) + B(P-P_u):
    A(P_s-P) + B(P-P_u)
    =AP_s-AP + BP-BP_u
    =(B-A)P+(AP_s-BP_u)
    Which has to equal 1. That means that B-A=0 and AP_s-BP_u=1. This is two linear equations in the two unknowns A and B, so you can solve for them. The answer ends up being pretty intuitive - not always the case with partial fractions. I usually like to double-check my final partial fraction result before moving on.

    There's another way to figure out what A and B are, which is based on the principle that if two polynomials in P are equal, then they have to be equal for all values of P.

    - Hollywood
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