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Help with differential equation

  1. Aug 5, 2008 #1
    I've been having trouble with this one differential equation for a VERY long time. This is not homework.

    [tex]\frac{dx}{y+z}[/tex] = [tex]\frac{dy}{x+z}[/tex] = [tex]\frac{dz}{x+y}[/tex]

    Any suggestions on where to start? Any advice will be much appreciated.

    Last edited: Aug 5, 2008
  2. jcsd
  3. Aug 5, 2008 #2


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    Staff: Mentor

    Is it still coursework? If so, I need to move this to the Homework Help forums.
  4. Aug 5, 2008 #3
    No, it's in a book I'm using to learn how to solve differential equations; a century-old book.

    The book is here:


    On page 259, problem 3. The solution is with it.

    I just need to see how to arrive at that answer. It's been bugging me for a month or two. =\
    Last edited: Aug 5, 2008
  5. Aug 5, 2008 #4

    Since I can't quite browse the book (assuming such function should be available), could you possibly post here the solution provided? Furthermore, and I ask as I'm not very acquainted with the notation used, are you asked to solve for z = z(x), y = y(x)? If that is the case, how about y = z = x ?
  6. Aug 5, 2008 #5
    The solution is:

    [tex]\sqrt{x+y+z}[/tex] = [tex]\frac{a}{z-y}[/tex] = [tex]\frac{b}{x-z}[/tex]

    On account of the chapter, I do not think that x, y, and z are functions, more like the three interdependent variables.
  7. Aug 6, 2008 #6
    Hi there!

    I'm not sure if this would help solving the system, but I'll post it :)


    from it, we obtain the system:


    where [tex]y\rightarrow y(x)[/tex] and [tex]z\rightarrow z(x)[/tex]



    now, adding the two equations we get:

    [tex]y'+z'=\frac{x+z}{y+z}+\frac{x+y}{z+y}[/tex] or

    now we define [tex]u\rightarrow u(x)[/tex], such that [tex]u(x)=y(x)+z(x)[/tex]

    this means tnat: [tex]u'(x)=y'(x)+z'(x)[/tex], or briefly [tex]u'=y'+z'[/tex]

    so our equation gets reduced to:

    [tex]u'=\frac{2x+u}{u}[/tex], which is a first order nonlinear equation for [tex]u(x)[/tex]

    Now unfortuanltey, I have no idea how the last equation could be solved :(
  8. Aug 6, 2008 #7
    Here's also my desperate attempt to solve the remaining equation:



    [tex]\displaystyle{\int}uu' dx-\displaystyle{\int}u dx=\displaystyle{\int}2x dx[/tex]

    [tex]\displaystyle{\int}uu' dx=u^2-\displaystyle{\int}u'u dx[/tex]

    [tex]\displaystyle{2\int}uu' dx=u^2[/tex]

    and hence

    [tex]\displaystyle{\int}uu' dx=\frac{u^2}{2}[/tex]

    so plugging it in the above equation

    [tex]\frac{u^2}{2}-\displaystyle{\int}u dx=x^2[/tex]

    [tex]U(x)=\displaystyle{\int}u dx=\frac{u^2}{2}-x^2[/tex]

    which is I think an integral equation

    maybe someone can solve it (if it's solvable) and then resubstitute [tex]u=y+z[/tex] to obtain the solution to the system

    best regards, Marin
    Last edited: Aug 6, 2008
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