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Math Help - Method of Undet Coefficient on this DE?

  1. #1
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    Method of Undet Coefficient on this DE?

    Hey would you use the Method of Undetermined Coefficients (Method of Good Guessing) to find a particular sol'n to this DE

    y''-5y'+6y=7e^{2x}+cos(2x)
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  2. #2
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    And you would have to split the particular solution up because the non homogeneous part is the sum of two terms?

    In other words find Yp for y''-5y'+6=7e^{2x} then find Yp for y''-5y'+6=cos(2x) and the two of them together make up the particular solution for the original DE?
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  3. #3
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    Quote Originally Posted by mmattson07 View Post
    Hey would you use the Method of Undetermined Coefficients (Method of Good Guessing) to find a particular sol'n to this DE

    y''-5y'+6y=7e^{2x}+cos(2x)
    Solve the homogeneous DE:

    y'' - 5y' + 6y = 0

    Characteristic Equation:

    m^2 - 5m + 6 = 0

    (m - 2)(m - 3) = 0

    m = 2 or m = 3.


    So y_c = Ae^{2x} + Be^{3x}.


    Now, notice that we already have a term of the form e^{2x} in y_c. So all we need to worry about now in the particular solution is C_1\cos{2x} + C_2\sin{2x}.

    y_p = C_1\cos{2x} + C_2\sin{2x}

    y_p' = 2C_2\cos{2x} - 2C_1\sin{2x}

    y_p'' = -4C_2\sin{2x} - 4C_1\cos{2x}.


    Substitute these into your DE and solve for C_1 and C_2.

    Then y = y_c + y_p.
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  4. #4
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    EDIT: I substituted the appropriate Yp's and I got

    2C2(sin(2x))+2C1(cos(2x))-10C2(cos(2x))+10C1(sin(2x))=7e^2x + cos(2x).

    I'm still unsure how to determine C1 and C2 from this?
    Last edited by mmattson07; March 22nd 2010 at 09:27 PM.
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  5. #5
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    Ok so I set the coefficients equal and found C1= \frac{1}{52} and C2= \frac{-5}{52}. I believe I need to find a particular solution for the 7e^{2x} part in which id guess Yp to be Axe^{2x}
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    From the homogenous solution to the differential you find that solutions are:

    y=c_1e^{2x}+c_2e^{3x}.

    However, when you go to solve for the two particular solutions to the differential, you notice that you already have an exponential to the 2x as a solution of your homogenous. So you need to multiple your original "guess" by x^{k}, where k is the smallest exponent needed to get a unique guess that is not a solution to our homogenous differential.

    Therefore I would solve my particular as:

    y_p = Axe^{2x}
    y'_p = Ae^{2x}+2Axe^{2x} \Rightarrow Ae^{2x}[1+2x]
    y''_p=2Ae^{2x}[1+2x]+Ae^{2x}[2] \Rightarrow 4Axe^{2x}+4Ae^{2x}

    Now substituting into your original d.e.:

    4Axe^{2x}+4Ae^{2x}-5(Ae^{2x}+2Axe^{2x})+6(Axe^{2x})=7e^{2x} \Rightarrow
    4Axe^{2x}+4Ae^{2x}-5Ae^{2x}-10Axe^{2x}+6Axe^{2x}=7e^{2x}

    Terms cancel out leaving:

    -Ae^{2x}=7e^{2x}

    Therefore our particular solution for our exponential is -7xe^{2x}
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  7. #7
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    Thanks. I calculated the full solution to be

    \frac{-45}{4}(e^{2x})+\frac{120}{13}(e^{3x})+\frac{1}{52}  (cos(2x))-\frac{5}{52}(sin(2x))-7xe^{2x}
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