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Solve 6x^2*y^2*dx+4x^3*y*dy=0.

  1. Oct 10, 2014 #1
    1. The problem statement, all variables and given/known data
    Solve 6x^2*y^2*dx+4x^3*y*dy=0.

    2. Relevant equations
    This is an exact equation.

    3. The attempt at a solution
    Here's my work:

    6x^2*y^2*dx=-4x^3*y*dy
    1/x dx=-2/3*1/y*dy
    ln abs(x)+C=-2/3*ln abs(y)
    1/y^(2/3)=Cx
    y^(2/3)=1/(Cx)
    xy^(2/3)=1/C
    C=1/(xy^(2/3))
    But the answer in the book says 2x^3*y^2=C. Is the book's answer correct and my answer is wrong or vice versa? Please correct me and tell me what to do if I'm wrong.
     
  2. jcsd
  3. Oct 10, 2014 #2

    Mark44

    Staff: Mentor

    Your answer is equivalent to the books answer.
    Starting with this: xy2/3 = 1/C = C', cube both sides to get x3y2 = (C')2 = C''. Here the primes do not denote differentiation - just different constants with slightly different names. To get to the book's answer, multiply the last equation above by 2.

    Tip: if your solution differs from the "official" book solution, verify that your solution works by differentiating it to see if you can get back to the differential equation. If your solution satisfies the DE, you're golden.

    You can take it a step further, and verify that the book's solution also works. If not, it wouldn't be the first time that a published solution contains a typo or is otherwise wrong. If it works, you should be able to manipulate one equation into the other.
     
  4. Oct 10, 2014 #3
    But even though my answer is equivalent to the book's answer, I think I should work this problem out using the exact equation method. Do you know that method?
     
  5. Oct 10, 2014 #4

    Mark44

    Staff: Mentor

    Yes, I do.
    Assuming you have an equation of the form M(x)dx + N(y)dy = 0, if My = Nx, the equation is exact.
    To solve, integrate M(x) with respect to x. You should get a "constant" term that's actually a function of y alone, say g(y).
    Integrate N(y) with respect to y. You should get another "constant" term that is a function of x alone, say f(x).
    Reconcile the differences, and that's your solution.
     
  6. Oct 10, 2014 #5

    Mark44

    Staff: Mentor

    My post #4 deserves a bit of elaboration.
    Again assuming that your equation M(x)dx + N(y)dy = 0 is exact, here's what's going on.

    The goal is to get to the solution, in the form of F(x, y) = C
    From here, I'm going to work backwards to the differential equation.

    If we take the total differential of this equation, or d(F(x, y)) = d(0), we get Fx(x, y)dx + Fy(x, y)dy = 0. Here the subscripts denote the two first partials of F. What we're doing when we check for exactness is verifying that Fxy = Fyx. IOW, that the mixed partials are equal.

    Now we can see that M(x) in the equation at the top of this post is Fx(x, y) and N(y) is Fy(x, y).

    After we've determined that the equation is exact, we get back to the unknown function F (a function of x and y) by taking the "x" antiderivative of M(x) and the "y" antiderivative of N(y) to get F(x, y) = C, the solution.
     
  7. Oct 10, 2014 #6
    So I got...
    2x^3*y^2+g(y)
    2x^3*y^2+f(x)
    Do I set them equal to each other like this?
    2x^3*y^2+g(y)=2x^3*y^2+f(x)
     
  8. Oct 10, 2014 #7

    Mark44

    Staff: Mentor

    No.
    The first line is really F(x, y) = 2x^3*y^2+g(y).
    The second is really F(x, y) = 2x^3*y^2+f(x).
    The first line does not have any function of x alone, so f(x) in the second line must be zero. Likewise, since the second line contains no function of y alone, g(y) in the first line must also be zero.

    What you have is F(x, y) = C, so your solution is what?
     
  9. Oct 10, 2014 #8

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    @Mark44: I presume you already know it, but I always warned my students that although the heuristic method of taking the two anti-partial derivatives and figuring out the answer by examining them often works, that process is not well defined and can lead to incorrect answers.

    I posted an article about that in the old library, which is now in the FAQ section. The link is:
    https://www.physicsforums.com/threads/how-to-find-potential-functions.762892/
     
    Last edited: Oct 10, 2014
  10. Oct 11, 2014 #9
    Thank you, Mark44. I got the right answer.
     
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