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Homework Help: Checking if an equation is exact and finding the solution

  1. Feb 9, 2015 #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 F(x,y) whose differential, dF(xy) is the left hand side of the differential equation. That is, level curves F(xy)=C are solutions to the differential equation:


    My(xy)= , and Nx(xy)= .
    If the equation is not exact, enter not exact, otherwise enter in F(xy) =

    2. Relevant equations

    3. The attempt at a solution
    I partially differentiated 1xy2-2y and 1x2y−2x to get My and Nx respectively. Since they both turned out to be 2xy-2, I concluded that the equation was exact.

    At this point, I'm a little lost on what to do. I tried following my notes but I was sick that week in class so I didn't really put coherent/pertinent notes down. Put simply, I'm stuck and super confused. Some websites are telling me to integrate then differentiate, others are telling me just integrate. I'm not sure. I looked at the thread a few days ago that had a very similar problem but could not discern whether the method used was a general solution or a solution for just that problem. Thank you in advance!
    Last edited by a moderator: Feb 9, 2015
  2. jcsd
  3. Feb 9, 2015 #2


    Staff: Mentor

    You have M = xy2 - 2y and N = x2y - 2x
    Integrate M with respect to x, keeping in mind that the "constant" of integration is some unknown function of y alone.
    Integrate N with respect to y, this time keeping in mind that the "constant" of integration will be a function of x alone.
    Compare the two answers you got above, and adjust things so that the two answers reconcile -- what you get will be F(x, y), where M = Fx(x, y) and N = Fy(x, y).

    By determining that your equation (i.e., M dx + N dy = 0) is exact, what you're doing is finding a function F, in two variables, so that the left side of your differential equation is the total differential of F. Since dF = 0, it follows that F must be a constant. IOW, F(x, y) = 0.
  4. Feb 9, 2015 #3
    Thank you for your thorough response!

    Is my partial differentiation correct then?
    When I integrated I got
    M = x2y - 2x + h(y)
    N = xy2 - 2y + h(x)

    With h(y) and h(x) being those two constants of integration you mentioned.
    I'm confused on what you mean by "adjust things so that the two answers reconcile". Where do I go from here?

    Again, thank you for your help. It means a lot.
  5. Feb 9, 2015 #4


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    @Mark44 "Reconciling the answers" by comparing them. That's the rub. That method, while it often works, is unreliable especially in the hands of a student just learning the material. Look at the 2d example in:
  6. Feb 9, 2015 #5
    I'm sorry, that example confused me further, though I'm sure it is helpful.
    The method I have discerned from my notes is that I should partial differentiate the equation, to see if the equation is exact. Upon finding that it is, I need to integrate My(?) and then the rest of my notes aren't clear. Is there a clear method to solving these types of questions?
  7. Feb 9, 2015 #6


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    You didn't integrate correctly.
  8. Feb 9, 2015 #7


    Staff: Mentor

    ... on either one. In fact, I don't see that you did any integration.
  9. Feb 9, 2015 #8
    I integrated the new equation I got from partially differentiating. 2xy - 2.
    Is that not what I should have done?

    Okay let me take another look at it.
  10. Feb 9, 2015 #9


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    If you read the 2 variable example in the link I gave above, it shows you the proper method. I've got to run for now.
  11. Feb 9, 2015 #10
    @Mark44 I integrated the partially differentiated equation, not the ones you said were M and N. I mixed them up, sorry.
    Those integrated are
    M = (x2y2)/2 - 2xy + h(y)
    N = (x2y2)/2 - 2xy + h(x)

    They are the same. LCKurtz's link is a bit complicated and I fail to see how to apply it to this particular question.
  12. Feb 9, 2015 #11


    Staff: Mentor

    Your work is correct but those aren't M and N. Each one is a kind of "view" of F(x, y). Also, you should use different letters for what you're calling h(y) and h(x). They shouldn't both be h. Better might be g(x) and h(y).

    If N had ended up with some function of y, that would be what h(y) is. And if M had ended up with some function of x, that's what would be g(x). Since neither is present, you can conclude that both g(x) and h(y) actually are just constants.
  13. Feb 10, 2015 #12


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    You seem to be not quite clear on the reasons behind this method you are using or what those symbols mean.
    If F(x,y) is differentiable, and x and y are themselves differentiable functions or some parameter t, then, by the chain rule
    [tex]\frac{dF}{dt}= \frac{\partial F}{\partial x}\frac{dx}{dt}+ \frac{\partial F}{\partial y}\frac{dy}{dt}[/tex]
    Since the parameter t is not relevant here, we can write this in "differential form":
    [tex]dF= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy[/tex]

    Your test for "exactness" works because, with [itex]M= \frac{\partial F}{\partial x}[/itex] and [itex]\frac{\partial F}{\partial y}[/itex], [itex]N_x= \frac{\partial}{\partial x}\left(\frac{\partial F}{\partial y}\right)= \frac{\partial F^2}{\partial x\partial y}[/itex] and [itex]M_y= \frac{\partial}{\partial y}\left(\frac{\partial F}{\partial x}\right)= \frac{\partial F^2}{\partial y\partial x}[/itex] and those "mixed second derivatives" must be equal.

    In this problem, you have [itex]M= \frac{\partial F}{\partial x}= xy^2- 2y[/itex]. Since differentiating with respect to x treats y as a constant, we get F (not M) by integrating with respect to x, treating y as a constant: [itex]F= \frac{x^2y^2}{2}- 2xy+ f(y)[/itex] where "f(y)" is the "constant of integration. Since we are "treating y as a constant" that "constant of integration" may be a function or y. That is what you have but you had them equal to "M" and "N", not F. M and N are the partial derivatives of F.

    Now, do NOT integrate N! Instead, differentiate what we have for F with respect to y:
    [itex]F_y= x^2y- 2x+ f'(y)[/itex]. That f' really is an ordinary derivative since f is a function only of y.
    That must be equal to "N": [tex]F_y= x^2y- 2x+ f'= x^2y- 2x= N[/tex].
    All the terms that involve "x" cancel. That had to happen in order to leave f' as a function of y only and is the result of this being an exact differential. Here, it happens that every term involves "x" so we are left with f'= 0. That is f really is a costant so [itex]F= \frac{x^2y^2}{2}- 2xy+ C[/itex].

    Since the given equation said "dF= 0", F must be a constant. You get your solution by setting that last formula equal to a constant. (And, of course, you can combine the two constants.)

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