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First order nonlinear differential equation Help needed.

  1. Sep 7, 2011 #1
    Hi All,

    I have been trying to solve following nonlinear differential equation, but I couldn't solve it.

    0 = a*[f(t)]^{z/(z-1)} + (-t+C)*f(t) + b*[df(t)/dt]

    where a, b and C are constants and 0< z<1.

    Could you please help me how to solve this nonlinear differential equation? I would really appreciate any suggestions. Many thanks, in advance!

    Karen
     
  2. jcsd
  3. Sep 7, 2011 #2
  4. Sep 7, 2011 #3
    Thanks a lot for the suggestion, but unfortunately I could not make integrating factor technique work since integrating factors turns out to be a function of both "t" and "f". Any other suggestions?

    Kind regards
     
  5. Sep 7, 2011 #4
    I got a suggestion: When you get stuck, fall back and work on a simpler one first, then build it back up to the original problem:

    [tex]f'+f^{1/2}+(c-t)f=0[/tex]

    now, what happens if I let [itex]u=f^{1/2}[/itex]

    you can convert that to a DE in u right?. You know, f=u^2, f'=2uu'. You can do that? Ok, put all that back in, solve it for u, then of course f=u^2. Ok, how about:

    [tex]f'+f^{3/4}+(c-t)f=0[/tex]

    same dif. Let u=f^{1/4}. Now when you figure out what u is, then raise it to the fourth power to get f.

    How about:

    [tex]f'+f^{n/m}+(c-t)f=0[/tex]

    How about when n/m is not rational?
     
    Last edited: Sep 7, 2011
  6. Sep 7, 2011 #5
    I will definitely try to solve it in the way that you suggested jackmell. Thanks a lot!

    As a quick note, z/(z-1) is negative; hence it may cause some problems.
     
  7. Sep 7, 2011 #6
    As I said, z/(z-1) taking negative values is very problematic, therefore unfortunately, I could not solve the ODE using the approach you suggested, jackmall. I appreciate your time, though.

    Any other suggestions would be highly appreciated.

    Kind regards
     
  8. Sep 8, 2011 #7
    If:

    [tex]f'+\frac{1}{f^{1/2}}+(c-t)f=0[/tex]

    then what happens if we let:

    [tex]u=f^{3/2}[/tex]

    If:

    [tex]f'+\frac{1}{f^{1/4}}+(c-t)f=0[/tex]

    then what happens if we let:

    [tex]u=f^{5/4}[/tex]
     
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