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Is this ODE separable?

  1. Aug 8, 2005 #1
    I'm almost finished my calculus book (I'm self-teaching) and in the last 2 chapters it's giving a brief intro to differential equations. the second section is for "separable" and I'm stuck on this one halfway through the exercises. It doesn't seem to be separable by any means I can see unless there's some kind of substitution (which he's never mentioned anywhere yet).

    my book gives:
    [tex](y^2 - x^2)dy + 2xydx = 0[/tex]

    the closest I can seem to get it is (1):
    [tex]\frac{x^2-y^2}{2xy} = \frac{dx}{dy}[/tex] *or* [tex]\frac{2xy}{x^2-y^2} = \frac{dy}{dx}[/tex]

    or (2):
    [tex]\frac{1}{2}(\frac{x}{y}-\frac{y}{x}) = \frac{dx}{dy}[/tex]

    or even (3):
    [tex]y dy - \frac{x^2}{y}dy + 2xdx = 0[/tex]

    Now... I have the Schaum's "3000 solved problems in calculus" and in it there's a problem which simplifies into form (1) I have up there and goes on to say it's a "homogeneous" so substitute in y=vx.
    Is my problem even a separable one? Excuse my DE newbieness. :yuck:
  2. jcsd
  3. Aug 8, 2005 #2


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    This would make more sense in a broader view in which one would recognize it as homogeneous.
    you could solve
    then realize the sum of the two solutions is the solution of the original problem.
    or you could realize
    d(x/y) and d(y/x) should be involved, then try to obtain the differential equation in terms of them.
    Last edited: Aug 9, 2005
  4. Aug 9, 2005 #3


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    yes, it's homogenous, one way to test is to substitute ax and ay for every x and y, and see if the constants cancel out to give you the original form. In pertinence to the "schaum's" example you mentioned....

    [tex]y=vx,~y'=v+v'x [/tex]

    [tex]\frac{2xy}{x^2-y^2} = \frac{dy}{dx}[/tex]

    [tex] \frac{2x^{2}v}{x^{2}(1-v^{2})} = \frac{dv}{dx}x+v[/tex]

    now try separating
  5. Aug 9, 2005 #4
    thanks guys. very strange that my book would do this. he doesn't go over homogeneous for another couple of sections but he goes and gives me a problem on it. at least now that I know what it is I can go about figuring it out.

    but anyway...
    what is the reasoning behind the substitution of y=vx? and how do you get y' = v + v'x from that?
  6. Aug 10, 2005 #5


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    The fact that function is "homogeneous" in x and y really means that it can be written as a function of y/x. For example, [tex]\frac{2xy}{x^2-y^2}[/tex] is homogeneous because it has the same "power" of the variables (2) in both numerator and denominator. If you divide both numerator and denominator by x2, you get [tex]\frac{2\frac{y}{x}}{1- (\frac{y}{x})^2}[/tex]. Let v=y/x and that becomes [tex]\frac{2v}{1-v^2}[/tex].
    Taking v= y/x is the same as y= vx. And, of course, if y= vx then y'= v(x)'+ v'x by the product rule. And that is y'= v+ v'x since (x)'= 1.
  7. Aug 10, 2005 #6
    ahh. thank you HallsofIvy. it makes sense to me now.
  8. Feb 1, 2010 #7
    Just in case, there is a test for separability for any ODE in the form dy/dx = F(x,y). In other words, can we write F(x,y) as f(x)g(y). You can google this test.
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