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Did I solve this DE correctly?

  1. Dec 10, 2006 #1
    If someone has a chance out there, could you please check my math here and let me know if I am doing this correctly or not.

    Problem:

    Solve: (2x-1)dx + (3y+7)dy = 0

    I would like to solve this using the "Exact" method for solving DE's, so:

    [tex]\frac{\partial_P}{\partial_y}(2x-1) = 0[/tex]
    [tex]\frac{\partial_Q}{\partial_x}(3y+7) = 0[/tex]

    [tex]\int{2x-1dx} = x^2-x + g(y)[/tex]

    [tex]\frac{\partial_F}{\partial_y}x^2-x+g(y) = g'(y)[/tex]

    [tex]g'(y) = 3y+7 -> g(y) = \frac{3}{2}y^2+7y[/tex]

    [tex]F(x,y) = x^2-x+\frac{3}{2}y^2+7y[/tex]

    How does that look?
     
    Last edited: Dec 10, 2006
  2. jcsd
  3. Dec 10, 2006 #2
    Also, one quick question. Does the equation have to be in the form of P(x.y)dx + Q(x,y)dy = 0? Can it be minus instead of plus? The reason I ask is because I vaugly remember hearing something about that at the begining of the school quarter and can't seem to find it in my notes now.

    Thanks!
     
  4. Dec 11, 2006 #3

    dextercioby

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    Incomplete.

    [tex] x^2-x+\frac{3}{2}y^2+7y=C [/tex]

    would be better.

    Daniel.
     
  5. Dec 11, 2006 #4

    dextercioby

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    The minus can always be absorbed into Q, so it's always a "+". Kinda simple of you think about, right? :bugeye:

    Daniel.

    EDIT: Wow, i hit the "back" button and this reply got into another thread...
     
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