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Math Help - Assignment: Differential Equation

  1. #1
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    Question Assignment: Differential Equation

    I have this one question for my Mathematical Modelling unit Assignment 1 which I have been stuck on for hour 8+ hours (deadly serious 0.0).

    dP/dt = (1/2)*P(4-P)) P(0) = 2

    I try solving it as a separable DE and I get to;

    P/(4-P) = A*e^(2t) (Starting to think that this part might be wrong too -.-)

    From this point I have tried everything I know to try and solve for P(t)

    I REALLYYYYYYY need some help with this one.
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  2. #2
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    Re: Assignment: Differential Equation

    \displaystyle \begin{align*} \\ \\ \frac{P}{4 - P}\ &=\ \frac{P - 4 + 4}{4 - P} \\ \\ &=\ \frac{4 - (4 - P)}{4 - P} \\ \\ &=\ \frac{4}{4 - P} - 1 \end{align*}
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  3. #3
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    Re: Assignment: Differential Equation

    Hello, stripe501!

    \frac{dP}{dt} \:=\:\tfrac{1}{2}P(4-P) \qquad P(0) = 2

    We have: . \frac{dP}{dt} \:=\:-\tfrac{1}{2}P(P-4)

    Separate variables: . \frac{dP}{P(P-4)} \:=\:-\tfrac{1}{2}\,dt

    Partial fractions: . \tfrac{1}{4}\left[\frac{1}{P-4} - \frac{1}{P}\right]dP \:=\:-\tfrac{1}{2}\,dt

    Integrate: . \tfrac{1}{4}\int\left(\frac{1}{P-4} - \frac{1}{P}\right)\,dP \:=\:-\tfrac{1}{2}\int dt

    . . . . . . . . . \tfrac{1}{4}\bigg(\ln|P-4| - \ln|P|\bigg) \:=\:-\tfrac{1}{2}t + c_1

    Multiply by 4: . . \ln|P-4| - \ln|P| \:=\:-2t + c_2

    . . . . . . . . . . . . . . . . . \ln\left|\frac{P-4}{P}\right| \:=\:-2t + c_2

    Exponentiate: . . . . . . . . . \frac{P-4}{P} \:=\:e^{-2t+c_2} \:=\:e^{-2t}\cdot e^{c_2}

    . . . . . . . . . . . . . . . . . . . \frac{P-4}{P} \:=\:Ce^{-2t}

    . . . . . . . . . . . . . . . . . . . P - 4 \:=\:CPe^{-2t}

    . . . . . . . . . . . . . . . . P - CPe^{-2t} \:=\:4

    . . . . . . . . . . . . . . . P(1 - Ce^{-2t}) \:=\:4

    . . . . . . . . . . . . . . . . . . . . . . P \:=\:\frac{4}{1-Ce^{-2t}} .[1]


    We are given: P(0) = 2 . . . when t = 0,\:P = 2

    . . 2 \:=\:\frac{4}{1-Ce^0} \quad\Rightarrow\quad 2 \:=\:\frac{4}{1-C} \quad\Rightarrow\quad 2 - 2C \:=\:4

    . . -2C \:=\:2 \quad\Rightarrow\quad C \:=\:-1


    Substitute into [1]: . \boxed{P \;=\;\frac{4}{1 + e^{-2t}}}
    Thanks from stripe501
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  4. #4
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    Re: Assignment: Differential Equation

    Thanks so much , you made a few mistakes in the working but that part that I was stuck on had the same method to solve. I got the answer to be P= 4e^(2t)/(1+e^(2t)). Thanks for your help!
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