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Analitical solution for dy/dt=ay^3+by^2+cy+d

  1. Jul 25, 2012 #1
    Dear all

    I have to solve an equation which seems to be simple:

    It is dy/dt=ay^3+by^2+cy+d but my problem is that I am not able to give the analitical solution y(t)=... And I need your help for that...

    What I have already done is :

    - I know that you have to find the solutions of ay^3+by^2+cy+d=0 (the coefficients allow me to say that three solutions exist and lead to y1, y2 and y3 as real solutions)

    - Then one can write :
    dy/(ay^3+by^2+cy+d)=edy/(y-y1)+fdy/(y-y2)+gdy/(y-y3), with e, f, g easy to determine.

    - Then you have to integrate on both sides of the equality : edy/(y-y1)+fdy/(y-y2)+gdy/(y-y3)=-dt

    - which leads to something like that [(y-y1)^e][(y-y2)^f][(y-y3)^g]=exp(-t)

    --> MY PROBLEM IS : do you know a way to express directly the analitical solution : y(t)=...? using this formulae, or by using a different integration method
     
  2. jcsd
  3. Jul 25, 2012 #2

    LCKurtz

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    I think you will get ##e^{+t}## on the right. To get an explicit solution instead of an implicit one, you need to solve a cubic equation for ##y##. That can be done but except for special cases, it isn't pretty.
     
  4. Jul 25, 2012 #3
    I think it's worst than that. He has:

    [tex](y-y_1)^a(y-y_2)^b(y-y_3)^c=e^t[/tex]

    No way you're getting a y out of there unless you're desperate and plot the data then invert the data and fit it to a nice-size polynomial.
     
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