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Solve For Exact Differential Equation

  1. Jan 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Solve the exact equation y^3-(14x+2)dx+3xy^2dy=0

    2. Relevant equations

    NA

    3. The attempt at a solution

    I proved these were exact because dM/dy and DN/dx both equal 3y^2

    I chose to work with N first and df/dy=3xy^2

    Therefore f(x,y)=xy^3+h(x)

    I took df/dx of this and got y^3 + h'(x)

    I made this equal to the other df/dx so it looks like:
    df/dx = y^3+h'(x)=y^3-(14x+2)

    h'(x)=-14x+2 and so h(x) is -7x^2+2x (+ constant)

    I plugged this into the original problem with h(x) so it now looks like:

    f(x,y)=xy^3-7x^2+2x=C

    Solving for y, I get y=(-7x^2+2x+C)^(1/3) / x^(1/3)

    This is not correct and I don't know what I'm missing here.
     
  2. jcsd
  3. Jan 31, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You have a basic algebra error! -(14x+ 2)= -14x- 2, not -14x+ 2.
     
  4. Jan 31, 2012 #3
    Ah yes. I did it again from the beginning and was able to solve it correctly. Thanks for pointing out the error. :)
     
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