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Nonnlinear Differential Equation problem

  1. Jul 15, 2005 #1
    Nonlinear Differential Equation problem

    Hi all,

    I'm new here. I stumbled upon the site while searching for solutions to nonlinear differential solution.

    I have a problem that i've been cracking my head to solve for the past 2 days to no avail. I hope you guys can help me out here. I'm looking for a general analytical solution.

    Here's the problem:

    y'' + Ay^2 = B , where A and B are constants

    Help will be greatly appreciated.

    Thank you all in advance.
    Last edited: Jul 15, 2005
  2. jcsd
  3. Jul 15, 2005 #2


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    Oh sorry, i am tired, i will come back to it later :zzz:.

    Good luck
    Last edited: Jul 15, 2005
  4. Jul 15, 2005 #3
    But the problem is in the second order. Can we still use the simple seperation of variable method?
  5. Jul 15, 2005 #4
    this is one of those "standard trick" problems.

    introduce a new variable v(x) = dy/dx;
    then the equation is

    dv/dx + Ay^2 = B

    now use the chain rule: dv/dx = (dv/dy) (dy/dx) = v dv/dy

    now you have an equation: v dv/dy = B-Ay^2
    which is separable. so you integrate it, and
    then replace v with dy/dx again and hopefully
    solve it from there.
  6. Jul 15, 2005 #5
    I've already tried that method qbert and i ended up something's that more complicated, which is as follow:

    upon resolving:

    v dv/dy = B - Ay^2

    I got the following expression when i substituted the original expression of v(x) = dy/dx back into the solution:

    dy/dx = [2(By-Ay^3/3)]^0.5

    I was stuck from then onwards.
  7. Jul 15, 2005 #6
    All right. I think i've sold the first order differential equation of dy/dx = [2(By-Ay^3/3)]^0.5.

    It can be solved by introducing a new variable of z = y^0.5 which will result in the expression:

    y = (3B/A)*sin[(B/2)^0.5*x]

    Thanks anyway for all the help guys!
  8. Jul 15, 2005 #7


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    edit: don't use the bernoulli, forgot this was a second order equation

    use chain rule

    [tex] v=y^{'},~\frac{dv}{dt}= \frac{dy}{dt} \frac{dv}{dy} [/tex]

    [tex]v \frac{dv}{dy} = \frac{d^{2}y}{dt}= \frac{dv}{dt} [/tex]

    [tex]v \frac{dv}{dy} +Ay^{2}=B [/tex]

    [tex] v \frac{dv}{dy}=B-Ay^{2} [/tex]

    the above is separable
    Last edited: Jul 15, 2005
  9. Jul 15, 2005 #8


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    well, seems that you've already got to this point, I'm not quite sure what you (kank_39) were referring to in z=y^.5, can you explain?
  10. Jul 15, 2005 #9
    it would be a better solution if it solved the problem.

    I don't mean to be a jerk. but, you did this step wrong. The constants of
    integration can't be dropped if you're looking for the general solution.
    When you get done you should have two linearly independent solutions
    with appropriate integration constants.

    you should end up with
    [tex] \frac{dy}{dx} = \pm \sqrt{ 2\left(By - \frac{Ay^3}{3}\right) + c_1 } [/tex]
    which is seperable

    [tex] \int dx = \int \frac{\pm}{\sqrt{ 2\left(By - \frac{Ay^3}{3}\right) + c_1} } dy + c_2 [/tex]

    This integral doesnt have a "nice" solution (unless A, B, and c_1 are "special"),
    you can rewrite it in terms of elliptic integrals if that makes you feel better.
  11. Jul 18, 2005 #10
    z = y^0.5 represents z to be equal to the square root of y or y to the power of 0.5. :)
  12. Jul 18, 2005 #11
    Oh.. sorry, i forgot to mention the initial conditions which is v(y=0) = y(x=0) = 0. That would allow [tex]c_1[/tex] to become zero and the results to be as i've solved it.

    However, i'm now trying to solve this:

    [tex] \int dx = \int \frac{\pm}{\sqrt{ 2\left(By - \frac{Ay^3}{3}\right) + c_1} } dy + c_2 [/tex]

    for the same problem but with a different set of initial condition and i'm trying to understand the concept of elliptic integrals as i've never encountered such a theory before.
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