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Exact ODE

  1. Sep 25, 2012 #1
    1. The problem statement, all variables and given/known data

    From Elementary Differential Equation by Boyce and Diprima
    Chapter 2 Miscellaneous Problems #11
    (x^2+y)dx + (x+e^x)dy = 0

    ANS:(x^3/3)+xy+e^x=c

    2. Relevant equations

    multiplying an integrating factor to make the DE exact:
    1. du/dx = u(My - Nx)/ N

    2. du/dx = u(Nx-My)/ M

    3. The attempt at a solution

    First try: I guessed this can be changed into exact DE so, I tried with the two above equation:
    equation 1 gave me:

    du/u = e^x/(x+e^x)
    I don't know how to solve this...

    then equation 2 gave me:

    u = e^((e^x)*ln(x^2+y))

    I am not sure if multiply this integrating factor to the original DE will make it exact...

    Second try: I manipulated the given DE and changed it to a linear form:

    dy/dx = -(x^2+y)/(x+e^x)

    dy/dx + 1/(x+e^x) * y = (-x^2)/(x+e^x)

    and I found integrating factor to be:

    I = e^∫1/(x+e^x) dx

    which I am unable to solve...
     
  2. jcsd
  3. Sep 26, 2012 #2
    Third try:

    take integral of both side

    ∫(x^2+y)dx + (x+e^x)dy = ∫0

    (x^3)/3 + yx + xy + ye^x = c //move constant c from left side to right side

    (x^3)/3 + 2xy + ye^x = c

    however it's not quite the same as the answer...

    Wolframalfa does not solve this one!!!
     
  4. Sep 26, 2012 #3
    ## (x^3/3)+xy+e^x=c ## cannot be the solution for ##(x^2+y)dx + (x+e^x)dy = 0##.

    Differentiate: [tex]
    d((x^3/3)+xy+e^x) = d(x^3/3) + d(xy) + d(e^x)
    \\ = x^2dx + xdy + ydx + e^xdx
    \\ = (x^2 + y + e^x)dx + xdy \ne (x^2+y)dx + (x+e^x)dy
    [/tex]
     
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