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Solving a first order DE using integrating factor method

  1. Feb 13, 2012 #1
    1. The problem statement, all variables and given/known data
    (3(x^2)y + 2xy + y^3)dx + (x^2 + y^2)dy = 0


    3. The attempt at a solution
    This is from my notes, so I already have the answer. I just don't understand the very last step with the integrations.

    (3(x^2)y + 2xy + y^3)dx + (x^2 + y^2)dy = 0

    My = 3x^2 + 2x + 3y^2
    Nx = 2x

    (My-Nx)/N = (3x^2 + 3y^2)/(x^2 + y^2) = 3
    dlnu/dx = 3, u = e^(3x)

    Multiply by integrating factor u = e^(3x)
    (3(x^2)y + 2xy + y^3)e^(3x)dx + (x^2 + y^2)e^(3x)dy = 0

    and then the part I don't understand:
    phi(x,y) = [∫ 0 to y (x^2 + y^2)(e^(3x))dy] + [∫ 0 to x (0 dx)]
    = (x^2)y(e^(3x)) + (1/3)(y^3)(e^(3x))

    Thanks
     
  2. jcsd
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