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Math Help - Non-homogeneous Equations Initial Values Problem

  1. #1
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    Non-homogeneous Equations Initial Values Problem

    Here is the problem:

    y−12y+32y=21ex

    y(0)=27 y(0)=21 y(0)=25

    Y(x) = ??

    I used the method of undetermined coefficients and my answer is:

    (71/3)+(8/3)e^(4x)+(-1/3)e^(8x)+e^x

    But the online-homework wouldn't accept it and says it's incorrect...

    Here is what I did:

    Yp(x) = (C1) + (C2)e^(4x) + (C3)e^(8x) + e^x
    Y'p(x) = 4(C2)e^(4x) + 8(C3)e^(8x) + e^x
    Y''p(x) = 8(C2)e^(4x)+64(C3)e^(8x) + e^x

    then after plug in and used the given initial values I got:

    (C1) + (C2) + (C3) + 1 = 27
    4(C2) + 8(C3) + 1 = 21
    8(C2) + 64(C3) + 1 = 25

    so C1 = 71/3
    C2 = 8/3
    C3 = -1/3

    Please help!!!

    Thank You!!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Phyxius117 View Post
    Here is the problem:

    y−12y+32y=21ex

    y(0)=27 y(0)=21 y(0)=25

    Y(x) = ??

    I used the method of undetermined coefficients and my answer is:

    (71/3)+(8/3)e^(4x)+(-1/3)e^(8x)+e^x

    But the online-homework wouldn't accept it and says it's incorrect...

    Here is what I did:

    Yp(x) = (C1) + (C2)e^(4x) + (C3)e^(8x) + e^x
    Y'p(x) = 4(C2)e^(4x) + 8(C3)e^(8x) + e^x
    Y''p(x) = 8(C2)e^(4x)+64(C3)e^(8x) + e^x (that 8 should be 16)

    then after plug in and used the given initial values I got:

    (C1) + (C2) + (C3) + 1 = 27
    4(C2) + 8(C3) + 1 = 21
    8(C2) + 64(C3) + 1 = 25 (again, the 8 should be 16)

    so C1 = 71/3
    C2 = 8/3
    C3 = -1/3 (the coefficients in the complementary solution should be different now.)

    Please help!!!

    Thank You!!
    Note the corrections in red.
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  3. #3
    Junior Member
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    Thank you so much!!! Just got the answer correct
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