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Math Help - Solve a linear differential equation

  1. #1
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    Solve a linear differential equation

    Hello, first time poster!
    I need help to solve a linear differential equation explicitly.
    Apparently this is a classic population-equation xn+1-xn=r(K-xn) where n is the generation, K is the supporting capacity (the maximum number of animals in a given enviroment).

    Given: K=240, x0=130, r=0,25

    I did solve this as recursion(?) in excel and I got some values which seemed okey.

    When I tried to solve this my general solution was Ce2x and my particular solution was 130e2x ,however, the values are not the same when I'm testing it to my recursive values. I tried to transform it to the form y'+ay=0. Does anyone have an idea what to do here?

    thanks in advance!
    Lenny
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  2. #2
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    Re: Solve a linear differential equation

    What you have is NOT a differential equation, it is a "finite difference" equation. Also "130e^{2x}" makes no sense because x is the function you are trying to solve for an the independent variable is n. Your solution has to be a formula for x_n as a funcition of n. You are correct that "linear finite difference equations with constant coefficients" are very similar to "linear differential equations with constant coefficients". The crucial difference is that while, if y= e^{ax} then y'= ae^{ax}= ay, with we use the fact that if x_n= a^{n}, x_{n+1}= a^{n+1}= a(a^n}= ax_n. So look for a solution of the form "Ca^n" rather than "Ce^{an}".
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  3. #3
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    Re: Solve a linear differential equation

    Quote Originally Posted by HallsofIvy View Post
    What you have is NOT a differential equation, it is a "finite difference" equation. Also "130e^{2x}" makes no sense because x is the function you are trying to solve for an the independent variable is n. Your solution has to be a formula for x_n as a funcition of n. You are correct that "linear finite difference equations with constant coefficients" are very similar to "linear differential equations with constant coefficients". The crucial difference is that while, if y= e^{ax} then y'= ae^{ax}= ay, with we use the fact that if x_n= a^{n}, x_{n+1}= a^{n+1}= a(a^n}= ax_n. So look for a solution of the form "Ca^n" rather than "Ce^{an}".
    I'm not quiet sure if I understand, will it be 130a^n? What is a?
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