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Math Help - Little help: Particular Integral for this equation

  1. #1
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    Little help: Particular Integral for this equation

    Im solving:

    y''-2y'+y= (2e^x)(cosx)

    The complimentary function after solving the auxiliary equation is:

    (A+Bx)e^x


    How should I start finding out the particular integral?

    by substituting

    y= k(e^x)(cosx) ( where only k is a constant)

    or y= k(e^p.x)(cos q.x) ( where k, p and q are constants)
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  2. #2
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    Are you familiar with Variation of Parameters?.

    y''-2y'+y=2e^{x}cos(x)

    m^{2}-2m+1=0\Rightarrow m=1

    Because 1 has multiplicity 2, we have:

    y_{c}=C_{1}e^{x}+C_{2}xe^{x}

    Note that y_{1}=e^{x}, \;\ y_{2}=xe^{x}

    Wronskians:

    W=\begin{vmatrix}e^{x}&xe^{x}\\e^{x}&(x+1)e^{x}\en  d{vmatrix}=e^{2x}

    W_{1}=\begin{vmatrix}0&xe^{x}\\2e^{x}cos(x)&(x+1)e  ^{x}\end{vmatrix}=-2xcos(x)e^{2x}

    W_{2}=\begin{vmatrix}e^{x}&0\\e^{x}&2e^{x}cos(x)\e  nd{vmatrix}=2e^{x}cos(x)

    {u'}_{1}=\frac{W_{1}}{W}=-2xcos(x)

    {u'}_{2}=\frac{W}{W_{2}}=2cos(x)

    Integrate the last two and get:

    u_{1}=-2(cos(x)+xsin(x))

    u_{2}=2sin(x)

    u_{1}+u_{2}=-2e^{x}cos(x)

    So, put them together with y_{c} and get

    \boxed{y=C_{1}e^{x}+C_{2}xe^{x}-2e^{x}cos(x)}
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  3. #3
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    Can you point me to some notes/slide on the Wronskians method.
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  4. #4
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    Do you have a DE book?. It will be in there under Variation of Parameters. Just google it. It's easy. It's just derivatives and determinants.
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