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Math Help - Solving a nonhomogenous D.E.

  1. #1
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    Solving a nonhomogenous D.E.

    (2x+y+1)y' = 1

    Although this intially appeared to be a simple problem I've tried every method I know how to do on this problem and I just can't get a solution.
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  2. #2
    MHF Contributor Calculus26's Avatar
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    dy/dx = 1/(2x+y+1)

    Invert:

    dx/dy = 2x + 2y +1

    dx/dy -2x = 2y +1

    use integrating factor e^(-2y)

    d[xe^(-2y)] = [(2y +1)e^(-2y)]dy

    x e^(-2y) = -e^(-2y)(y+1) + C

    x(y) = -(y + 1) + C e^(2y)
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  3. #3
    MHF Contributor Calculus26's Avatar
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    alternatively

    (2x+y+1)y' = 1


    can be written

    (2x+y+1)dy = dx

    -dx + (2x+y+1)dy = 0

    can be made exact with the integrating factor e^(-2y)

    -e^(-2y)dx + e^(-2y)(2x+y+1)dy = 0

    dF/dx = -e^(-2y)

    F = -xe^(-2y) + f(y)

    dF/dy = 2xe^(-2y) + f ' (y) = e^(-2y)(2x+y+1)


    f ' (y) = (y+1)e^(-2y)

    f(y) = -e^(-2y)(y+1)


    F = -xe^(-2y) -e^(-2y)(y+1) = C

    -x = Ce^(2y) +(y+1)

    x = -(y+1) + Ce^(2y)

    as in the previous post
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  4. #4
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    Thanks! My problem was when I integrated the I.F. I kept on integrating with respect to x instead of y. So I was getting e^(-2x) instead of e^(-2y).
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