1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Exact Differential Equations

  1. Oct 1, 2009 #1
    1. The problem statement, all variables and given/known data
    (28y6 - 24e-6y)(sqrt(1-t4))dy -(10t)dt =0

    Come up with a solution to this equation in the form of a function F(t,y)=0.


    2. Relevant equations
    dF/ dy = N
    dF / dt = M

    Note: when I write d it's supposed to be a partial derivative.
    3. The attempt at a solution
    Okay, so I called the part in front of the dy 'N' and the part in front of the dt 'M'. Then I took dM/dy and dN/dt to make sure they're equal. They are, so this equation is exact.

    F(t,y) = integral of (-10t)dt = -5t2 + k(y)

    Now, take the partial derivative of F(t,y) with respect to y. This gives -5t2y + k`(y). This equal to N.

    So -5t2y + k`(y) = (28y6 - 24e-6y)(sqrt(1-t4))

    There's where I get stuck. I need to figure out what k(y) is, but I have nooo idea how to do that. I tried moving everything over to the right side except k`(y) and integrating, but the integral got really complicated =\
     
  2. jcsd
  3. Oct 1, 2009 #2
    Remember that you have two variables so if you integrate with respect to one, you hold the other constant. For example if you integrate with respect to y, then you would treat t as a constant.
     
  4. Oct 1, 2009 #3
    Hmm, okay. I think I see where I went wrong.

    F = integral of M dt = -5t^2 + k(y)

    Now, take the partial derivative of F with respect to y. This gives k`(y).

    So k`(y) = (28y^6 - 24e^(-6y))(sqrt(1-t^4))

    So to find k(y), I integrated the right hand side of the above equation with respect to y. Then I got k(y) = (sqrt(1-t^4))(4y^7 + 4e^(-6y)).

    But my answer was still wrong...
     
    Last edited: Oct 2, 2009
  5. Oct 2, 2009 #4

    Mark44

    Staff: Mentor

    Why is it that you think this differential equation is exact? I'm not getting that at all. Based on your definitions for M and N,
    [tex]\frac{\partial M}{\partial y} = 0[/tex]
    and
    [tex]\frac{\partial N}{\partial t} = (28y^6 - 24e^{-6y})(1/2)(1 - t^4)^{-1/2}(-4t^3)[/tex]
    Are you sure you have written the problem down correctly?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Exact Differential Equations
Loading...