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Homework Help: Differential Equation Help

  1. Dec 6, 2006 #1
    1. The problem statement, all variables and given/known data
    Solve the following differntial equation

    2. Relevant equations
    (x+y-4)dy - (2x-y+1)dy = 0

    EDIT: Oops, it should be (x+y-4)dy - (2x-y+1)dx = 0

    3. The attempt at a solution
    I integrated both sides and ended up with:
    [y(2x +y-8)]/2 = X^2 + x(1-y) + c

    Did I solve this right, or am I supposed to do something else?
    Last edited: Dec 7, 2006
  2. jcsd
  3. Dec 6, 2006 #2


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    First of all: Which "dy" should be a "dx" instead??

    Secondly, you have probably done this incorrectly by fallaciously regarding "x" and "y" as constants in the integration on both sides. This is wrong.
  4. Dec 6, 2006 #3
    Did you mean [tex] \left( {x + y - 4} \right)dy - \left( {2x - y + 1} \right)dx = 0 [/tex]? If so then you have neglected some products.

    On the left, [tex] \int {xdy} \ne xy [/tex] and similarly on the right.
  5. Dec 6, 2006 #4


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    If it is, in fact, (x+y-4)dy - (2x-y+1)dy = 0
    which is the same as (x+y-4)dy + (-2x+y-1)dy = 0, then it
    is an "exact" equation: (x+y-4)x= 1= (-2x+y-1)y.
    There exist a function F(x,y) such that dF= (x+y-4)dy - (2x-y+1)dy = 0 and so F(x,y) is a constant. Do you know how to find that F?
  6. Dec 6, 2006 #5
    Isn't it? It's off by a constant, but that's it unless I'm crazy.
  7. Dec 7, 2006 #6
    No, because x is not a constant. Consider d(xy). Because x is not a constant this is not just xdy.
  8. Dec 7, 2006 #7
    Thanks for everyones help. I'm still a bit confused though. Any tips would be very much appreciated.
  9. Dec 7, 2006 #8
    You are integrating with respect to y, however, so as it is written above that will be equal to xy+c, since when you integrate with respect to y you would treat x as a constant.
  10. Dec 7, 2006 #9
    But when you differentiate with respect to y, you get

    d/dy (xy) = x + y * dx/dy

    and x is not treated as a constant, so why should it be treated as a constant in integration?
  11. Dec 7, 2006 #10
    If you had differentiated correctly there would be no problem. You differentiated with respect to x even though you wrote it as you were going to differentiate with respect to y. If you differentiate xy with respect to y, you treat x as a constant because you are not differentiating with respect to it, hence the derivative of xy with respect to y is simply x.
  12. Dec 7, 2006 #11


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    You cannot simply treat x as a constant in the term (x+y-4)dy and integrate it wrt y, since it is part of the differential equation. An equation can only be solved like that if it is separable into the form f(x)dx = g(y)dy, in which case both sides can be integrated. However, this equation cannot be separated into that form.

    On how to solve this equation... I'm stumped!
  13. Dec 7, 2006 #12
    as aero suggested, consider d(xy)=xdy+ydx
  14. Dec 8, 2006 #13
    This is only true for partial differentiation. However, here we are dealing with normal differentiation and integration.

    By your logic, d/dy (x) = 0 because you would treat x like a constant. But this is plain wrong: d/dy (x) = dx/dy.
    Last edited: Dec 8, 2006
  15. Dec 8, 2006 #14


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    We have had 4 people pointing out that d_leet was wrong. Dao Tuat, have you looked at my response. Can you answer my question?

    This is a "separable" differential equation. There must exist a function F(x,y) such that [itex]\frac{\partial F}{\partial x}= y- 2x- 1[/itex] and [itex]\frac{\partial F}{\partial y}= x+ y- 4[/itex]. Can you find F?
    Last edited by a moderator: Dec 8, 2006
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