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Homework Help: Find value of K for differential equation to be exact

  1. Sep 27, 2015 #1
    1. The problem statement, all variables and given/known data
    Hello everyone,
    I need to find K for the following differential equation to be exact.
    (y3+kxy4-2*x)dx +(3xy2 +20x2y3)dy=0

    2. Relevant equations

    3. The attempt at a solution
    (y3+kxy4-2*x)dx +(3xy2 +20x2y3)dy=0

    dM/dy = d/dy(y3+kxy4-2*x) = 4*k*y3*x+3*y2

    dN/dx =d/dx(3xy2 +20x2y3) = 40*x*y3+3*y2

    After that I really have no idea how to continue this.

    thank you.
  2. jcsd
  3. Sep 27, 2015 #2
    You have done 98% of it. What value of k makes the two expressions equal?
  4. Sep 27, 2015 #3


    Staff: Mentor

    Your diff. equation will be exact if My = Nx; that is, if 4kxy3 + 3y2 = 40xy3 + 3y2. What does k have to be so that the two expressions are identically equal?
  5. Sep 27, 2015 #4
    oh wow I am so sleepy.

    K have to be 10.
    But in this particular equation its pretty easy to spot out, but what if both part is very different, how do I proceed?
  6. Sep 27, 2015 #5


    Staff: Mentor

    Yes, k = 10. If the equation isn't exact, there are several alternatives, but they are too involved to explain in a small space like this. If you read ahead in your book, they might have some discussion of these techniques.
  7. Sep 27, 2015 #6
    thanks also, to find the solution of current equation after finding k=10

    I do
    integral(40*x*y3+3*y2)dx = 20*x2*y3+3*x*y2
    integral(40*x*y3+3*y2)dy =10*y4*x+y3

    I look for same terms in both results, but here I see that there isnt.

    so c= 20*x2*y3+3*x*y2+ 10*y4*x+y3

    is that correct ?
  8. Sep 27, 2015 #7


    Staff: Mentor

    No. Integrate M with respect to x, keeping in mind that there might be a "constant" of integration that would be either a constant or a function of y alone, say g(y).
    Then integrate N with respect to y, also keeping in mind that there could be a "constant" of integration that would be a constant or a function of x alone, say h(x).

    What you're doing here is assuming that there is an equation F(x, y) = C. If you take the total derivative of both sides, you get
    $$\frac{\partial F}{\partial x} dx + \frac{\partial F}{\partial y} dy = 0$$
    Here, M = ##\frac{\partial F}{\partial x}## and N = ##\frac{\partial F}{\partial y}##, and you're trying to reconstruct F from its two partial derivatives.
  9. Sep 27, 2015 #8
    I don't understand.

    M(x,y) = 40*x*y3+3*y2
    N(x,y)= 40*x*y3+3*y2

    I don't simply integrate each ?
  10. Sep 27, 2015 #9


    Staff: Mentor

    From the original problem, M = y3 + 10xy4, and N = 3xy2 + 40xy3.

    What you have above are My and Nx. The condition for exactness is that these two partials should be equal. To solve the diff. eqn, you need to do as I suggest in my previous post.
  11. Sep 27, 2015 #10
    = integral(y3+10xy4-2*x)dx
    = x2*(5*y4-1)+x*y3 +h(y)
    = x2*5*y4-x2+x*y3 +h(y)
    h(y) = n/a

    integral(3xy2 +20x2y3)dy
    = 5*y4*x2+y3*x +g(x)
    g(x) = -x2

    c= x2*5*y4-x2+x*y3
  12. Sep 27, 2015 #11


    Staff: Mentor

    Yes, looks good.

    To check, take the total differential of both sides. (I mistakely said total derivative earlier.)

    Your solution F(x, y) = C, where ##F(x, y) = 5x^2y^4 + xy^3 - x^2##
    Taking the total differential of both sides, we get d(F(x, y)) = d(C)
    or, ##\frac{\partial F}{\partial x}dx + \frac{\partial F}{\partial y}dy = 0##
    With the partial derivatives calculated, you should get your original diff. equation.
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