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Math Help - Finding Non-Homgenous Differential Equations

  1. #1
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    Finding Non-Homgenous Differential Equations

    Hi

    I am having problem trying to find y_p of the equation:

    y=2\frac{d^2y}{dx^2}+4\frac{dy}{dx}+7y=e^{-x}cos(x)

    i tried y_p = Asin(x)+Bcos(x) but this doesn't work, because i am still left with a constant of e^{x}.


    P.S
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  2. #2
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    First of all, don't use equals signs where things aren't equal. y does NOT equal the rest of that stuff...

    I would try y_p = e^{-x}(A\cos{x} + B\sin{x}).


    Then \frac{dy_p}{dx} = -e^{-x}(A\cos{x} + B\sin{x}) + e^{-x}(-A\sin{x} + B\cos{x})

     = e^{-x}[-(A + B)\sin{x} + (B - A)\cos{x}]


    and \frac{d^2y_p}{dx^2} = -e^{-x}[-(A + B)\sin{x} + (B-A)\cos{x}] + e^{-x}[-(A + B)\cos{x} + (A-B)\sin{x}]

     = e^{-x}(2A\sin{x} - 2B\cos{x}).



    Substituting into your DE gives:

    2\frac{d^2y}{dx^2} + 4\frac{dy}{dx} + 7y = e^{-x}\cos{x}

    2e^{-x}(2A\sin{x} - 2B\cos{x}) + 4e^{-x}[-(A + B)\sin{x} + (B - A)\cos{x}] + 7e^{-x}(A\cos{x} + B\sin{x}) = e^{-x}\cos{x}

    e^{-x}(4A\sin{x} - 4B\cos{x}) + e^{-x}[-(4A + 4B)\sin{x} + (4B - 4A)\cos{x}] + e^{-x}(7A\cos{x} + 7B\sin{x}) = e^{-x}\cos{x}

    e^{-x}[4A\sin{x} - 4B\cos{x} - (4A + 4B)\sin{x} + (4B - 4A)\cos{x} + 7A\cos{x} + 7B\sin{x}] = e^{-x}\cos{x}

    e^{-x}[3B\sin{x} + 3A\cos{x}] = e^{-x}\cos{x}

    3B\sin{x} + 3A\cos{x} = 0\sin{x} + 1\cos{x}.

    Therefore 3B = 0 and 3A = 1

    B = 0 and A = \frac{1}{3}.


    This means y_p = \frac{1}{3}e^{-x}\cos{x}.
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  3. #3
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    well after looking at the equation i tried another way where you separate e^{-x} and cos(x) and do them separately, however i don't get the same answer as what you have done.
    This is my working solution

    y_{p1}= Ce^{-x}

    y_{p1}' = -Ce^{-x}

    y_{p1}'' = Ce^{-x}

    2(Ce^{-x}-Ce^{-x}-Ce^{-x})+ 4(-Cxe^{-x}+-Ce^{-x})+7(-Cxe^{-x})=-e^{-x}cos(x)

    2Cxe^{-x}-2Ce^{-x}-2Ce^{-x}+ -4Cxe^{-x}+4Ce^{-x}+7Cxe^{-x})=-e^{-x}cos(x)<br />

    9Cxe^{-x}-4Ce^{-x}=e^{-x}

    5C=1

    C=\frac{1}{5}

    y_{p2} = Asin(x)+Bcos(x)

    y_{p2}' = Acos(x)-Bsin(x)

    y_{p2}''= -Asin(x)-Bcos(x)

    2(-Asin(x)-Bcos(x))+4(Acos(x)-Bsin(x))+7(Asin(x)+Bcos(x))=cos(x)

    -2Asin(x)-2Bcos(x))+4Acos(x)-4Bsin(x))+7Asin(x)+7Bcos(x))=cos(x)

    sin(x)(5A-4B)+cos(x)(5B+4A)=cos(x)

    5A-4B=0

    5B-4A=1

    A=\frac{4B}{5}

    5B-4(\frac{4B}{5})=1

    \frac{25B}{5}+\frac{16B}{5}=1

    B=\frac{5}{41}

    A=4\frac{\frac{5}{41}}{5}
    A=\frac{4}{41}

    final answer i get is:

    y= e^{-x}(Acos(\sqrt{\frac{5}{2}}x)+ Asin(\sqrt{\frac{5}{2}}x)+\frac{1}{5}e^{-x}* \frac{4}{41}sin+\frac{5}{41}cos(x)
    what is my problem??
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  4. #4
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    Well you're obviously not going to get the same answer, because a sum of an exponential with a trigonometric function is not the same as a product of them.

    I don't know why you're overcomplicating this. The easiest way is to use the same "family" of functions as you are given.

    In this case, your "family" is the product of an exponential and the trigonometric functions.

    The only time you would have to do differently is if this "family" appears in your homogenous solution. Then you have to multiply it by x.
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  5. #5
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    ok i understand now, i think i was confusing myself.
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