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Exact Equations

  1. Jul 11, 2012 #1
    Function is

    (x - 2xy + e^y) dx + (y - x^2 + xe^y) dy = 0

    Okay so it is the standard convenient exact equation for newbies. Now here is the part that confuses me.

    (a) Let P(x,y) = (x - 2xy + e^y) dx & Q(x,y) = (y - x^2 + xe^y) dy

    The function is defined for all real Numbers on an x,y plane. So we start with the initial integration of x_i = 0 & y_i = 0

    Now the author reasons that the integration of Q(x,y) is simply equal to the integration of y

    (from 0 to y) ∫ (y - x^2 + xe^y) dy = (from o to y) ∫ y dy

    Okay so it makes sense, since the initial value of x is zero, and since we are keeping x constant, it therefore follows that x will be 0 for all values and we can simply leave it at that.

    Now the part that confuses me is that he does not do the same for the other function P(x,y) dx regardless of the fact that y_i = 0.

    (a) gives x^2/2 - x^2y + xe^y + y^2/2 = c
     
  2. jcsd
  3. Jul 11, 2012 #2

    Simon Bridge

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    It can help to understand the motivation by going through the process in reverse vis:
    http://tutorial.math.lamar.edu/Classes/DE/Exact.aspx

    aside:
    ... who? What does he do instead?

    Remember I cannot see what you are looking at, you have to tell me - or provide a link.
    I'm guessing that you mean that Q was integrated wrt y and P wrt x.... which leads to the suggestion above ;)
     
  4. Jul 11, 2012 #3

    HallsofIvy

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    Strictly speaking, "dx" and "dy" are not part of the P and Q. What you want to say is that
    df= P(x,y)dx+ Q(x,y)dy so that [itex]P(x,y)= x- 2xy+ e^y[/itex] and [itex]Q(x,y)= y- x^2+ xe^y[/itex].

    Well, I can't speak for "the author" since I don't quite follow that (I don't see where the "ydy" on the right comes from). But I would say that, if f(x,y) is a differentiable function of x and y then we have
    [tex]df= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}dy[/tex]
    so we must have
    [tex]\frac{\partial f}{\partial x}= x - 2xy + e^y[/tex]
    integrating with respect to x, [itex]f(x,y)= (1/2)x^2- x^2y+ xe^y+ \phi(y)[/itex].
    (Since the partial derivative with respect to x treats y as a constant, the "constant of integration" may be a function of y hence the "[itex]\phi(y)[/itex]".)

    Differentiating that with respect to y we get
    [tex]f_y(x,y)=-x^2+ xe^y+ \phi'(y)[/tex]
    where, because [itex]\phi[/itex] is a function of y only, that last derivative is an ordinary derivative.
    Looking back to the original equation we must have
    [tex]\frac{\partial f}{\partial y}= -x^2+ xe^y+ \phi'(y)= y - x^2 + xe^y[/tex]
    so that [itex]\phi'(y)= y[/itex] and [itex]\phi= (1/2)y^2+ C[/itex] where, since [itex]\phi[/itex] really is a function of the single variable, y, C, the "constant of integration, reall9y is a constant.

    Putting that into [itex]f(x,y)= (1/2)x^2- x^2y+ xe^y+ \phi(y)[/itex], we have [itex]f(x,y)= (1/2)x^2- x^2y+ xe^y+(1/2)y^2)[/itex]. Because the original differential equation was "df= 0", f is a constant: [itex](1/2)x^2- x^2y+ xe^y+(1/2)y^2)= C[/itex]
     
  5. Jul 12, 2012 #4
    Let me highlight the main points.
    (x - 2xy + e^y) dx + (y - x^2 + xe^y) dy = 0

    He sets the initial integration of x_i = 0 & y_i = 0

    Q(x,y) = (y - x^2 + xe^y)

    Quote from author in this example, "with x_i = 0, Q(x_i,y) = y"

    I know, exact equations are very straight forward to me now I've got them down. I picked up this book and used its exact equation chapter to review https://www.amazon.com/Ordinary-Dif...rds=elementary+differential+equations+pollard

    It is very good book and it is definitely my favorite in differential equations, but I just don't know the motivation behind what he did (stated in the original post).

    I'm sorry that didn't help, I understand exact equations for the most part. Just not what the author did in this case.
     
    Last edited: Jul 12, 2012
  6. Jul 14, 2012 #5
    Hey guys, I figured it out. I was mis-interpreting the proof and technique that he proposed. What happens is that in his technique, initial x in Q(x_0,y) is 0, so when you integrate with respect to y, you can cancel out all the terms that have x! It is a very nice technique so I included the proof (it also serves as the technique itself).

    Most of the words I copied out of the book, credit goes to the book by Pollard and Tenenbaum, called ordinary differential equations.

    A necessary & sufficient condition that the differential equation

    [itex] P(x,y) dx + Q(x,y)dy = 0[/itex] be exact is that [itex]\frac{∂}{∂y} P(x,y) = \frac{∂}{∂x} Q(x,y) [/itex]
    Where the functions exist and are continuous in a region R.

    Proof of sufficient condition

    The function f(x,y) if it exists, must have a property that [itex]/frac{∂ f(x,y)}{∂x} = P(x,y) [/itex]

    With y constant we obtain [itex] eq (1) f(x,y) = \int_{x_o}^{x} P(x,y) dx + R(y)[/itex]

    Where x_o is a constant & R(y) stands for the arbitrary constant of integration.

    By definition, function f(x,y) must have the property that [itex] \frac{∂}{∂y}f(x,y) = Q(x,y)[/itex]

    [itex] \frac{∂}{∂y} \int {x_o}^{x} P(x,y) dx + R'(y) = Q(x,y)[/itex]

    By hypotheses P(x,y) is continuous, and by a theorem in analysis we may do the following:

    [itex] \int {x_o}^{x} \frac{∂}{∂y} P(x,y) dx + R'(y) = Q(x,y)[/itex]

    Which is the same as

    [itex] \int_{x_o}^{x} \frac{∂}{∂x} Q(x,y) dx + R'(y) = Q(x,y) [/itex]

    [itex] \left[Q(x,y) \right]_{x_o}^{x} + R'(y) = Q(x,y) [/itex]

    Which simplfies to [itex] R'(y) = Q(x_o,y)[/itex]
    [itex]R(y) = \int_{y_o}^{y} Q(x_0,y) dy[/itex], where y_0 is a constant
    Substituting in eq 1 we get

    [itex] f(x,y) = \int_{x_o}^{x} P(x,y)dx + \int_{y_o}^{y} Q(x_o,y) dy[/itex] is the solution

    If anyone needs any clarification, please let me know.
     
    Last edited: Jul 14, 2012
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