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Second order nonlinear ODE. How to begin solving it?

  1. Oct 4, 2014 #1
    1. The problem statement, all variables and given/known data
    This is not the exact problem that I want to solve but I will use this as a guidance tool:
    ##y'' - (y')^2 + y^3 = 0##
    where y is the function of x

    2. The attempt at a solution

    I tried doing a substitution ##u(x) = y'(x)## which leads to
    ##u' - u^2 + y^3 = 0## where both u and y are functions of x so this does not work.

    I tried doing a substitution ##u(y(x)) = y'(x)## which leads to
    ##u_y'y' - u^2 + y^3 = 0## here I have the derivative of u with respect to y and the derivative of y w. r. t. x which I never really encountered in solving these equations (maybe it is the right way and I should somehow proceed?).

    Tried substitutions with ##(y')^2## which did not get me far.

    Methods like variation of parameters cannot be applied here since there's no independant variable function (at least to the original dif. equation).

    Any tips? For the original problem the initial conditions are given but I suppose they are not necessary to understand the method of approaching these kind of problems?
     
  2. jcsd
  3. Oct 5, 2014 #2

    ehild

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    Homework Helper
    Gold Member

    The fusion of your ideas considering y as independent variable and substituting u = (y')2 will certainly work.
    Setting y as independent variable,
    ##y'' = \frac {dy'}{dy} \frac {dy}{dx}=0.5 \frac {d(y')^2}{dx}
    ##

    Substituting u=(y')^2:

    ##0.5 \frac {du}{dx}-u+y^3=0##

    You got a first-order linear equation for u, which can be solved by the integrating factor method, for example. Knowing u(y), dy/dx=sqrt(u(y)). You need to integrate, but it is not sure the integral can be obtained in terms of elementary functions.

    ehild
     
  4. Oct 5, 2014 #3
    Thank you for the help. I now recall getting the same integrating factor once but I abandoned the solution. I get confused when there is no answer in elementary functions which is the case here.
     
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