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Math Help - solving D.E by undetermined coefficients.

  1. #1
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    solving D.E by undetermined coefficients.

    question: y'' + 2y' - 24y = 16 - (x+2)e^4x

    what i've done:

    yh : λ^2 + 2λ - 24 = 0
    (λ+6)(λ-4) = 0
    yh = c1e^(-6x) + c2e^4x

    Try:
    yp = (A1x)e^4x <--- this is ok because it doesn't appear in yh

    y'p= (4A1x)e^4x + A1e^4x

    y''p= (16A1x)e^4x + 4A1e^4x + 4A1e^4x = (16A1x)e^4x + 8A1e^4x

    subbing into the D.E gives...

    (16A1x)e^4x + 8A1e^4x + 2[(4A1x)e^4x + A1e^4x] - 24[(A1x)e^4x] = 16 - (x+2)e^4x

    so grouping like terms gives: xe^4x: 16A1 + 8A1 - 24A1= 0 which is ok
    e^4x: 8
    A1 + 2A1 = x+2 , A1= (x+2)/10
    but where does the constant(16) come from?
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  2. #2
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    Quote Originally Posted by linalg123 View Post
    question: y'' + 2y' - 24y = 16 - (x+2)e^4x

    what i've done:

    yh : λ^2 + 2λ - 24 = 0
    (λ+6)(λ-4) = 0
    yh = c1e^(-6x) + c2e^4x

    Try:
    yp = (A1x)e^4x <--- this is ok because it doesn't appear in yh

    y'p= (4A1x)e^4x + A1e^4x

    y''p= (16A1x)e^4x + 4A1e^4x + 4A1e^4x = (16A1x)e^4x + 8A1e^4x

    subbing into the D.E gives...

    (16A1x)e^4x + 8A1e^4x + 2[(4A1x)e^4x + A1e^4x] - 24[(A1x)e^4x] = 16 - (x+2)e^4x

    so grouping like terms gives: xe^4x: 16A1 + 8A1 - 24A1= 0 which is ok
    e^4x: 8
    A1 + 2A1 = x+2 , A1= (x+2)/10
    but where does the constant(16) come from?
    You y_h is correct, but the form of your particular solution is incorrect.

    y_p=Ae^{4x}+Bxe^{4x}+C

    but since e^{4x} is part of the homogeneous solution we need to multiply exponential part of the particular solution by x this gives

    y_p=Axe^{4x}+Bx^2e^{4x}+C
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