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Homework Help: Is this equation exact or not?

  1. Apr 15, 2010 #1
    1. The problem statement, all variables and given/known data

    (x + (2/y))dy + ydx = 0

    2. Relevant equations
    i am hoping someone could help me with how to approach all of these sorts of problems, i mean like step by step approach

    3. The attempt at a solution

    M(x,y) =? N(x,y) (this is what i need to know right?)

    so, to differentiate 1st term with respect to y , i treat any x terms as constants right?

    if that is right, then :
    d/dy (x+ (2/y))dy = 0 + -2 / (y^2)

    then differentiate other term with respect to x and treat all y terms as constants

    d/dx ydx = 0

    therefore equation not exact..

    BUT if i differentiate 1st term with respect x and 2nd term with respect to y, then
    d/dx (x+ (2/y)) = 1


    d/dy ydy = 1

    so it is exact.

    how do i know which is correct? please help thanks
  2. jcsd
  3. Apr 15, 2010 #2


    Staff: Mentor

    Your equation is exact. To determine this you are trying to find a function f(x, y) such that fx(x, y) = M(x, y) = y, and fy(x, y) = N(x, y) = x + 2/y.

    Your equation is exact if
    [tex]\frac{\partial M(x, y)}{\partial y} = \frac{\partial N(x, y)}{\partial x} [/tex]

    Once you determine that your equation is exact, the goal is to find f(x, y) by 1) integrating M(x, y) with respect to x, 2) integrating N(x, y) with respect to y, 3) comparing the results.

    If the equation is exact, that what you have is d(f(x, y)) = 0 ==> f(x, y) = C. Here d(...) represents the total derivative, which is
    [tex]\frac{\partial f(x, y)}{\partial x} dx + \frac{\partial f(x, y)}{\partial y} dy[/tex]
  4. Apr 15, 2010 #3
    (x + (2/y))dy + ydx = 0

    M_x = N_y = 1. Therefore exact.

    integrate M(x,y)_y = -2/(y^2) + c
    integrate N(x,y)_x = 0 + c
  5. Apr 15, 2010 #4


    Staff: Mentor

    M(x, y) = y and N(x, y) = x + 2/y

    When you integrate a function of two variables with respect to one of the variables, instead of getting a constant of integration, you get a function that is considered to be a constant as far as differentiation is concerned.

    I'm having a hard time find the words to say that right, so here's an example:
    [itex]\int xy dx = (1/2)x^2y + f(y)[/itex]
    As a check, differentiate the right side with respect to x, and you get xy. The derivative with respect to x of a function of y alone is 0.
  6. Apr 15, 2010 #5
    I want to rework this since (i think) im understanding the whole process a little better..

    (x + (2/y))dy + ydx = 0

    y + (x + 2/y) dy/dx = 0


    DM_y = 1 = DN_x

    equation is exact

    INTEG-M_x: (1/2) x^2 + (2/y)x + h(y)

    D of INTEG-M_y: -(2x/y^2) + h'(y)

    h'(y) = x + (2/y) + (2x/(y^2))

    h(y) = ....this is where i am stuck.. Do i take derivative with respect to y or x or both and why? I am guessing only y , because its h(y)??

    if so, h(y) = 2ln|y|-4xy^-3

    and final answer would be:

    f(x,y) = -(2x/y^2) + 2ln|y|-4xy^-3

    am i doing this correctly? thanks for any help or tips
  7. Apr 15, 2010 #6


    Staff: Mentor

    Why don't you leave it like the above? There's no advantage in dividing by dx.
    No. If you integrate y wrt x, you get xy + f(x)

    Your next step is to integrate N wrt y. The work below here doesn't make any sense.
    The final answer is going to be F(x, y) = C, for some function F.

    When you get it, you can check by taking the total derivative, dF, getting dF = 0.

    [tex]dF = \frac{\partial F}{\partial x} dx + \frac{\partial F}{\partial y}dy[/tex]
    Or dF = Fx dx + Fy dy
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