Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Hi! how to solve? y(y-2x(y'))^3=(y')^2

  1. Mar 11, 2009 #1
    Hello Everybody! Have any suggestions how to solve this differential equation, I haven't.. :(
    y(y-2x(y'))^3=(y')^2
     
  2. jcsd
  3. Mar 11, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The only thing I can suggest is that you first treat it as a cubic equation for y' and try to solve for y'. But that obviously is going to be difficult!
     
  4. Mar 11, 2009 #3
    I tried this way, but it didn't give me any positive result. Moreover even Maple can't solve it exact.. it gives me huge expression..
     
  5. Mar 11, 2009 #4
    Are numeric or series solutions okay?
     
  6. Mar 11, 2009 #5
    The right solutions are y^2=2*(C^3)*x + C^2 and 27*(x^2)*(y^2)=1. Hmm.. Lagrange method doesn't fit properly here
     
  7. Mar 12, 2009 #6
    Hello math_addict,

    This equation can be solved by the following method. If a DE of the first order but not first degree can be written as [itex]x-\phi(y,y')=0[/itex] the way to solve it is by taking the derivative of it and consider y and y' as variables. Then you need to substitute dy/y' for dx and you get a first degree DE in y and y'. After integrating this equation you need to eliminate y' between this solution and the original equation to obtain the general solution. In this process you need to make sure that any factors that are eliminated are to be looked upon with care afterwards because they can be singular solutions. Let's try this on your problem.

    OK, first rewriting the original DE gives:

    [tex]x=\frac{y}{2y'}-\frac{1}{2y^{1/3}y'^{1/3}}[/tex]

    The derivative of this is:

    [tex]dx=\frac{1}{2}\left[\frac{1}{y'}+\frac{1}{3y'^{1/3}y^{4/3}}\right]dy+
    \frac{1}{2}\left[-\frac{y}{y'^2}+\frac{1}{3y^{1/3}y'^{4/3}}\right]dy'[/tex]

    Substituting [itex]\frac{dy}{y'}=dx[/itex] gives after rewriting:

    [tex]\frac{1}{2y'}\left[1-\frac{y'^{2/3}}{3y^{4/3}}\right]dy=
    -\frac{y}{2y'^2}\left[1-\frac{y'^{2/3}}{3y^{4/3}}\right]dy'[/tex]

    The factor between brackets can be eliminated but is important for later on to see if it is a singular solution. We have thus so far:

    [tex]\frac{dy'}{y'}=-\frac{dy}{y}[/tex]

    Which has as solution:

    [tex]y'=\frac{C}{y}[/tex]

    Substituting this in the original equation gives you:

    [tex]y\left(y-2x\frac{C}{y}\right)^3=\frac{C^2}{y^2}[/tex]

    Or, when we set C=K^3:

    [tex]y^2=K^2+2xK^3[/tex]

    This is the general solution to the DE. Now the factor that was eliminated needs to be looked upon. Setting it equal to zero:

    [tex]1-\frac{y'^{2/3}}{3y^{4/3}}=0[/tex]

    Or, after rewriting:

    [tex]y'=3\sqrt{3}y^2[/tex]

    Which has the solution:

    [tex]-\frac{1}{y}=3\sqrt{3}x[/tex]

    Or:

    [tex]27x^2y^2=1[/tex]

    And it is not possible to get this by setting K to any number and thus is a singular solution because it is also a solution to the original DE.

    Hope this helps, coomast
     
  8. Mar 13, 2009 #7
    Thanks, Coomast, for help.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Hi! how to solve? y(y-2x(y'))^3=(y')^2
  1. How to solve y''=0 (Replies: 1)

Loading...