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: Exact DE's

  1. Feb 5, 2006 #1

    Pengwuino

    User Avatar
    Gold Member

    How do you determine a solution to a DE if you know the DE is exact? My professor (or possibly my notes) did a horrible job of explaining how to do it and the book does an even worse job explaining how to do it. For example, how would i go about solving this equation? I need to find c, the constant.

    [tex] ( - 2xy^2 + 2y)dx + ( - 2x^2 y + 2x)dy = 0[/tex]

    I know that I differentiate each part by dx/dy and dy/dx to attain:

    [tex]- 2xy + 2\frac{{dx}}{{dy}} = - 2xy + 2\frac{{dy}}{{dx}}[/tex]

    But where do I go from that point?
     
  2. jcsd
  3. Feb 5, 2006 #2
    first you took the derivative wrong it should be -4xy+2

    after that you would intergate and combine terms

    can you do this
     
  4. Feb 5, 2006 #3

    Pengwuino

    User Avatar
    Gold Member

    haha oops oh yah... missed that twice it seems.

    Which part do i integrate?
     
  5. Feb 6, 2006 #4
    you would integrate the original functions M and N

    M = -2xy^2 + 2y dx

    N = -2x^2y +2x dy
     
  6. Feb 6, 2006 #5

    Pengwuino

    User Avatar
    Gold Member

    Ok they both integrate to be:

    -x^2y^2+2xy

    now what?
     
  7. Feb 6, 2006 #6
    [tex] ( - 2xy^2 + 2y)dx + ( - 2x^2 y + 2x)dy = 0[/tex]

    [tex] (-x^2y^2 + 2yx + c_1) + ( - x^2y^2 + 2xy + c_2 ) = 0 [/tex]

    [tex] -2x^2y^2 + 4yx + k = 0 [/tex]

    Edit: OOPS should be this:

    [tex] k = 2x^2y^2 -4xy[/tex]

    [tex] k= 2xy(xy - 2) [/tex]


    I have no clue, Im just playing along here. Don't take what I post seriously. Do you have an idea of what the solution should look like? I dont see how you can get a specific value of c without any boundary conditions.
     
    Last edited: Feb 6, 2006
  8. Feb 6, 2006 #7

    HallsofIvy

    User Avatar
    Science Advisor

    Your original equation was
    [tex] ( - 2xy^2 + 2y)dx + ( - 2x^2 y + 2x)dy = 0[/tex]
    That is an exact equation because (-2xy2+ 2y)y= -4xy+ 2 and (-2x2y+ 2x)x= -2xy+ 2 also.

    What that means is that there exist a function F(x,y) such that
    [tex]dF= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy= (-2xy^2+ 2y)dx+ (-2x^2y+ 2x)dx[/tex]
    (Above, we were checking the equality of the mixed second derivatives,
    [tex]\frac{\partial^2 F}{\partial x\partial y}= \frac{\partial^2 F}{\partial y\partial x}[/tex])

    That is, we must have
    [tex]\frac{\partial F}{\partial x}= -2xy^2+ 2x[/tex]
    Integrate that, with respect to x since in a partial derivative you are treating y as a constant. Of course, the "constant of integration" might, in fact, be a function of y. Once you get that expression for F, differentiate with respect to y and set it equal to
    [tex]\frac{\partial F}{\partial y}= -2x^2y+ 2y[/tex] to determine what that "constant" must be.
    In terms of that F, the equation just says "dF= 0" so F= constant is the general solution.
     
  9. Feb 7, 2006 #8

    Pengwuino

    User Avatar
    Gold Member

    So I set [tex] - x^2 y^2 + x^2 + c = - 2x^2 y + 2y[/tex] ??
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook