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Homework Help: No idea to do this differential equation queation.

  1. Sep 7, 2011 #1
    1. The problem statement, all variables and given/known data

    x2y1+xyy1-6y2=0
    find solution for this differential equation.

    y1 mean dy/dx

    How to do this question. I have no idea.
     
  2. jcsd
  3. Sep 7, 2011 #2

    Stephen Tashi

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    [tex] x^2 \frac{dy}{dx} + xy\frac{dy}{dx} - 6y^2 = 0 [/tex]

    I don't know how to do it systematically, but one may reason about it.

    If we assume [itex] y [/itex] is an nth degree polynomial in x, the degrees of the 3 terms are n+1, 2n, 2n. For the terms to cancel, we need n+1 = 2n, so n = 1.

    So try letting [itex] y = Cx + D [/itex] and solve for [itex]C[/itex] and [itex]D[/itex].

    If there is a "proper" way of doing it, I hope someone tells us.
     
  4. Sep 7, 2011 #3
    Hint: Try differentiating x^2*y^2.
     
  5. Sep 7, 2011 #4

    Ray Vickson

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    Maple 14 gets the solution as
    y = (x/5)*[1+Z(z)^5], where Z(x) is a solution of the equation 5z^6 - cxz^5 - cx=0, and c is an arbitrary constant. When you take c = 0 the solution is pretty simple; otherwise, it is likely horrible.

    RGV
     
  6. Sep 7, 2011 #5

    Stephen Tashi

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    I wonder if human beings are supposed to do it by "integrating factors"

    [tex] (x^2 + xy) \frac{dy}{dx} - 6y^2 = 0 [/tex]

    Is not an "exact" differential equation, but perhaps there is an integrating factor M(x,y) so that:
    [itex] M(x,y)(x^2 + xy)\frac{dy}{dx} - 6M(x,y)y^2 = 0 [/itex] is exact.
     
  7. Sep 7, 2011 #6
    sorry sorry sorry this is wrong question.
    The correct question.correct one is
    x2y12+xyy1-6y2=0
     
    Last edited: Sep 7, 2011
  8. Sep 7, 2011 #7

    SammyS

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    Factor this.

    Let u = x·y',

    then u2 + uy -6y2 = 0

    Factor the left hand side.

    (u+3y)(u-2y)=0

    Solve for u, then put x·y' back in for u .

    Can you take it from there?
     
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