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General Solution Differential Equation (integrating factors)

  1. Nov 5, 2011 #1
    1. The problem statement, all variables and given/known data

    find the general solution using integrating factors.
    xy' - 2y = 3x

    2. Relevant equations



    3. The attempt at a solution

    x(dy/dx) = 3x + 2y
    x*dy = (3x + 2y)dx
    (-3x + 2y)dx + xdy = 0

    My = -3x + 2
    Nx = 1
    Not Exact (hence the use of integrating factors)

    μ(x)(-3x + 2y)dx + μ(x)xdy = 0

    Differentiating with respect to y for M and x for N
    μ(x)(-3x + 2) = μ'(x)x + μ(x)

    trying to simplify
    -3xμ(x) + 2μ(x) - μ(x) = μ'(x)
    -3xμ(x) - μ(x) = μ'(x)

    -μ(x)(3x - 1) = μ'(x)

    I know I need to solve for μ(x) to multiply M and N by, I am not sure if I have done everything correct up until now. I assume that I possibly just need a little help with the algebra to be able to set up an integral to solve for μ(x) but I am not sure.

    Any help would be greatly appreciated, I'm not sure if I missed something from the beginning or am almost there. Thanks ahead of time!
     
    Last edited: Nov 5, 2011
  2. jcsd
  3. Nov 5, 2011 #2
    Just realized I did My wrong, re-trying now.
     
  4. Nov 5, 2011 #3
    Second attempt:

    Equations:
    F = e∫P(x)dx
    P(x) = 1/N*(My - Nx)
    ψx = M ψy = N

    Attempt:
    xy' - 2y = 3x
    -2y - 3x + xy' = 0
    My = -2 Nx = 1
    1/N(My - Nx)
    1/x(-2 - 1)
    3x-1
    ∫3x-1dx → -3ln|x|
    e-3ln|x| = -3x

    integrating factor = -3x

    (-3x)(-2y - 3x) + (-3x)xy' = 0
    (-6xy - 9x2) - (3x2)y' = 0
    My = -6x Ny = -6x EXACT

    Integrating M to get ψ(x,y)
    ∫ψxdx = ∫-6xy - 9x2dx
    ψ(x,y) = -3x2y - 3x3 + f(y)

    Differentiate with respect to y to get N
    ψy = -3x2 + f'(y)
    where N = ψy
    -3x2 = -3x2 + f'(y)
    0 = f'(y) → ∫f'(y)dy = ∫dx
    f(y) = c

    ψ = -3x2y - 3x3 + c

    I think I did everything right. I'm not sure if this is the correct answer or not any help or checking of my work would be appreciated thanks.
     
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