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

Homework Help: DE Question

  1. Sep 17, 2011 #1
    The given problem for multiple differential equations (not in a system) is: "Determine a region of the xy-plane for which the given differential equation would have a unique solution whose graph passes through a point (x0,y0) in the region."

    I'm not entirely sure what to do with this. Looking at the function:


    I found the unknown function to be:
    And I have verified that it is a solution.

    Now the books says the answer is: "half-planes defined by either y>0 or y<0"
    I'm not sure how this solution works. Could someone please explain it to me?

    I also don't understand why y=0 isn't valid. I can see y=0 satisfying this equation:

  2. jcsd
  3. Sep 18, 2011 #2


    User Avatar
    Science Advisor
    Homework Helper

    Yes, y=0 is a solution that passes through (0,0). y=(x/3)^3 is also a solution that passes through (0,0). So there is no unique solution if the initial point (x0,y0)=(0,0). What about (x0,y0) being a point besides (0,0)? I thinks that's what they are asking.
  4. Sep 18, 2011 #3


    User Avatar
    Homework Helper

    the DE is separable
    [tex] \frac{dy}{dx}=y^{\frac{2}{3}} [/tex]

    integrating both sides gives
    [tex] \int dyy^{-\frac{2}{3}}=\int dx [/tex]
    [tex] \frac{1}{3}y^{\frac{1}{3}}=x+c [/tex]
    [tex] \frac{1}{3}y^{\frac{1}{3}}=(\frac{x+c}{3})^3 [/tex]

    this is a family of solutions given by c

    now consider whether the following is a solution:
    [itex]for \ \ \ \ x \leq 0, \ \ \ \ \ \ y(x) = (\frac{x}{3})^3 [/itex]
    [itex]for \ \ 0< x \leq 1, \ \ y(x) = 0 [/itex]
    [itex]for \ \ \ \ 1< x, \ \ \ \ \ \ y(x) = (\frac{x-1}{3})^3 [/itex]
  5. Sep 18, 2011 #4
    So you are only using solutions which do not make the equation equal to zero? Is the term homogenous applicable to that case?
  6. Sep 18, 2011 #5


    User Avatar
    Science Advisor

    On the contrary, Lancelot59's solution explicitely includes y= 0. That's why he can construct an infinite number of solutions such that y(0)= 0.

    The "fundamental existence and uniqueness theorem for first order differential equations" says that the problem dy/dx= f(x,y), y(x0)= y0 has a unique solution if there is some neighborhood about (x0,y0) such that f(x,y) is continuous and f(x,y) is "Lipschitz" in y ("differentiable with respect to y is a simpler, sufficient but not necessary condition) in that neighborhood.

    Here, [itex]f(x,y)= y^{2/3}[/itex] is continuous in any neighborhood of (0,0) but [itex]f_y= (2/3)y^{-1/3}[/itex] does not exist at (0, 0) so the function is not Lipschtz there.

    We can guarentee a unique solution on any neighborhood that does NOT include the x-axis but not in any neighborhood that does.
  7. Sep 18, 2011 #6
    I see, so the trick is to differentiate the original differential equation with respect to the "dependent" (Y in this case) variable, and see where solutions for it are valid?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook