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Exact Differential Equation

  1. Jan 28, 2012 #1
    1. The problem statement, all variables and given/known data
    Use the "mixed partials" check to see if the following differential equation is exact.
    If it is exact find a function [itex]F(x,y)[/itex] whose differential, [itex]dF(x,y)[/itex] is the left hand side of the differential equation. That is, level curves [itex]F(x,y) = C[/itex] are solutions to the differential equation:
    [itex](-4xy^2+y)dx+(-4x^y+x)dy = 0[/itex]

    Find:
    [itex]M_{y}(x,y) = ?[/itex]
    [itex]N_{x}(x,y) = ?[/itex]
    [itex]F(x,y) = ?[/itex]


    2. The attempt at a solution
    I first partially differentiated each part (M and N) with respect to y and x, respectively. This got me this and they are exact:
    [itex]M(x,y) = -4xy^2+y[/itex]

    [itex]\frac{∂M}{∂y} = -8xy+1[/itex]

    [itex]N(x,y) = 4x^2y+x[/itex]

    [itex]\frac{∂N}{∂y} = -8xy+1[/itex]

    Then I integrated [itex]M(x,y)[/itex] with respect to x which got me:
    [itex]-2x^2y^2+xy+h(y)[/itex]

    Then I partially differentiated that equation with respect to y:
    [itex]\frac{∂M}{∂y} = -4x^2y+x+h'(y)[/itex]

    Next, I set that equation equal to [itex]N(x,y)[/itex] and solved for [itex]h'(y)[/itex]:
    [itex]-4x^2y+x+h'(y) = 4x^2y+x[/itex]
    [itex]h'(y) = 0[/itex]

    I then integrated with respect to y:
    [itex]∫ h'(y) dy = ∫ 0 dy[/itex]
    [itex]h(y) = 0 + C[/itex]

    Finally I plug it back into the original equation:

    [itex]-2x^2y^2+xy+h(y)[/itex]
    [itex]h(y) = 0 + C[/itex]
    [itex]-2x^2y^2 + xy + C = -2x^2y^2 + C[/itex]

    But because the website (WeBWorK) does not allow the input of "C", I just left it as:
    [itex]-2x^2y^2 + xy[/itex]

    Thus, my answers were:
    [itex]M_{y}(x,y) = -8xy+1[/itex]
    [itex]N_{x}(x,y) = -8xy+1[/itex]
    [itex]F(x,y) = -2x^2y^2 + xy[/itex]

    I submitted the answers but the site said they were wrong, and I don't know which ones are wrong because the question is one of those, "get all right for a point" kind of thing.
     
  2. jcsd
  3. Jan 28, 2012 #2

    LCKurtz

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    But what is the left side of that equation? What is it that you have just calculated? Remember that you are looking for a function F(x,y) such that Fx=M and Fy= N.
    What equation?? You have just differentiated an expression. You need an equals sign and something on both sides.
    And once you answer the above questions you might understand what is equal to that and also see why you should stop right there.

    Also, you can check whether your F(x,y) works (it does).
     
    Last edited: Jan 28, 2012
  4. Jan 28, 2012 #3
    My bad, the left side of that equation is suppose to be:
    [itex]∫ \frac{∂M}{∂y} dx[/itex]

    Which formulates this:
    [itex]∫ \frac{∂M}{∂y} dx = ∫ (-4xy^2+y) dx + h(y)[/itex]
    [itex]∫ \frac{∂M}{∂y} dx = -2x^2y^2 + xy + h(y)[/itex]

    I differentiated this:
    [itex]∫ \frac{∂M}{∂y} dx[/itex]

    Which led to this:
    [itex]\frac{∂}{∂y} [∫ \frac{∂M}{∂y} dx] = \frac{∂}{∂y} [-2x^2y^2 + xy + h(y)][/itex]
    [itex]\frac{∂}{∂y} [∫ \frac{∂M}{∂y} dx] = -4x^2y+x+h'(y)[/itex]
     
    Last edited: Jan 28, 2012
  5. Jan 28, 2012 #4

    LCKurtz

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    I added a note to my first post while you were typing this. Your answer is correct but your writeup isn't good. I know you were taking the antiderivative of M(x,y) with respect to x when you got ##-2x^2y^2+xy+h(y)##. But why are you doing that integration? What is it that you are calculating when you do that antiderivative? You need to understand that.
     
  6. Jan 28, 2012 #5
    Whops quick note, I just finished editing my previous post when you just responded; the equation you quoted from me should have been "dx" and not "dy".
     
  7. Jan 28, 2012 #6

    LCKurtz

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    Yes, I figured that...
     
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