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Homework Help: Differential equation

  1. Feb 16, 2008 #1
    [SOLVED] differential equation

    1. The problem statement, all variables and given/known data

    [tex]\frac{\partial y}{\partial x}=(x+y)^{2}, y(0)=0 [/tex]

    3. The attempt at a solution

    I have made two attempts, both using the same substitution, where I think I made an error.


    [tex] u=x+y, \partial u = \partial y, [/tex]
    [tex] \frac{\partial u}{\partial x}=u^{2}, [/tex]
    [tex] -\frac{1}{u}=x+c_{0}, [/tex]
    [tex] y=\frac{x^{2}+c_{0}+1}{x}, y(0)=0, [/tex]
    [tex] y=-x [/tex]

    checking the solution gives -1=0, which is false.


    [tex] u=x+y, \partial u = \partial y, [/tex]
    [tex] \frac{\partial u}{\partial x}=1, [/tex]
    [tex] (x+y)=x+c_{0}, [/tex]
    [tex] y=c_{0}=0 [/tex],

    which gives 0=x^{2} after putting it back in the differential equation.

    I'm sure I did something fundamentally wrong, and I don't know why I would be allowed to make these substitutions. I did them because every other way of solving this led to something I didn't believe. Could anyone enlighten me on the idea of substitution?
    Last edited: Feb 16, 2008
  2. jcsd
  3. Feb 16, 2008 #2
    The substitution [itex]u(x)=y(x)+x[/itex] gives [itex]u'(x)=1+y'(x)[/itex], thus your ODE reads [itex]u'(x)=u(x)^2+1[/itex] which is a separable one. Can you solve this one?
  4. Feb 16, 2008 #3
    what does the O in ODE mean?

    and how would you seperate your variables then? because u is dependent of x, I don't know what you're allowed to do next.
  5. Feb 16, 2008 #4


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    You may write:
  6. Feb 16, 2008 #5
    Ordinary Differential Equation = ODE. :smile:
    Just write
    [tex]u'=u^2+1\Rightarrow \frac{d\,u}{u^2+1}=d\,x[/tex]
    and integrate it.
  7. Feb 16, 2008 #6
    Ouppps! arildno was faster! :smile:
  8. Feb 16, 2008 #7
    okay I can do that, so thanks a lot :)

    but why are you allowed to integrate, because when u is dependent of x, you haven't really made seperated variables have you?
  9. Feb 16, 2008 #8


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    Okay; here you are mixing together the "separation of variables" technique in partial differential equations, and the chain rule trick in one-variable diff. eqs called "separation of variables".

    In partial diff.eqs, the "separation" has nothing to do with the chain rule of differentiation, but by ASSUMING a solution u(x,y)=F(x)*G(Y), i.e, that a solution u can be found by writing it as a product of two single-variable functions.
    Last edited: Feb 16, 2008
  10. Feb 16, 2008 #9
    you're right. I'm mistaken

    and thanks. I solved the problem (well, you helped too)... ;)
  11. Feb 16, 2008 #10


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    Don't use "curly d"s, these are not partial derivatives. In particular, x and y are not independent variables. The whole point of the equation is that y is a function of x!
    If u= x+ y, then du/dx= 1+ dy/dx.
    Using that, your equation becomes du/dx- 1= u2 so solve du/dx= u2+ 1.

  12. Feb 16, 2008 #11
    wow, you're all a nice help =)... duely noted and appreciated.
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