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Math Help - diff eq help

  1. #1
    suki0918
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    diff eq help

    Find the particular solution of the differential equation: y" - 4y' + 4y = e^(2x)?

    I'm not sure what to do b/c I get 0 for A?? Can someone help, this is what I had so far:

    Y(x) = Ae^2x
    Y'(x) = 2Ae^2x
    Y"(x) = 4Ae^2x

    So... 4Ae^2x - 8Ae^2x + 4Ae^2x = e^2x
    Which becomes 0 = e^2x

    and then I'm stumped...
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  2. #2
    Eater of Worlds
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    Have you ever used Variation of Parameters?.

    m^{2}-4m+4=0, \;\ (m-2)^{2}=0

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

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

    W=\begin{vmatrix}e^{2x}&xe^{2x}\\2e^{2x}&2xe^{2x}+  e^{2x}\end{vmatrix}=e^{4x}

    W_{1}=\begin{vmatrix}0&xe^{2x}\\e^{2x}&2xe^{2x}+e^  {2x}\end{vmatrix}=-xe^{4x}

    W_{2}=\begin{vmatrix}e^{2x}&0\\2e^{2x}&e^{2x}\end{  vmatrix}=e^{4x}

    u'_{1}=\frac{W_{1}}{W}=-x

    u'_{2}=\frac{W_{2}}{W}=1

    \int{-x}dx=\frac{-x^{2}}{2}

    \int{1}dx=x

    y_{p}=(\frac{-x^{2}}{2})e^{2x}+(x)xe^{2x}=\frac{x^{2}e^{2x}}{2}

    y=y_{c}+y_{p}=C_{1}e^{2x}+C_{2}xe^{2x}+\frac{x^{2}  e^{2x}}{2}

    y=\left(\frac{x^{2}}{2}+C_{1}+C_{2}x\right)e^{2x}

    Whew, it's time for beddy-bye
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  3. #3
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    Quote Originally Posted by suki0918 View Post
    Find the particular solution of the differential equation: y" - 4y' + 4y = e^(2x)?
    Galactus used Variation of Parameters.

    Another method is to use Method of Undetermined Coefficients.

    The general solution is y=C_1 e^{2x}+C_2xe^{2x}

    Since the RHS has e^{2x} we look for a solution of the form y=Ae^{2x}. But that is among the solutions! So we multiply by x to look for a solution of the form y=Axe^{2x}. But that is also among the solutions! So we look for a solution of the form y=Ax^2e^{2x}.
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  4. #4
    Newbie servantes135's Avatar
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    TPH,
    Why didn't you use "Thou shall not covet anothers work as thine own"?
    Lol verry funny though I must tell that to my Catholic friends.
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