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Math Help - the Wronskian and variation of parameters...

  1. #1
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    the Wronskian and variation of parameters...

    Consider L(y)=e^{2x}

    the Wronskian is W = \begin{vmatrix}<br />
e^{x} &  e^{2x}&e^{3x} \\ <br />
 e^{x}& 2e^{2x} & 3e^{2x}\\ <br />
 e^{x}&4e^{2x}  & 9e^{3x}<br />
\end{vmatrix}=2e^{6x}.

    ok so now I am supposed to write an annihilator for L(y)=0,
    then write the complementary solution y_{c},
    write an annihilator for L(y)=e^{2x}
    rewrite L(y)=e^{2x} in its expanded (y,y',...)
    notation.

    so the annihilator is (D-1)(D-2)(D-3)
    y_{c}=C_{1}e^{x}+C_{2}e^{2x}+C_{3}e^{3x}

    the last two parts of the problem im not sure how to answer, and to be honest im not sure if the work i have done up to this point is correct. any help would be appreciated.
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  2. #2
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    Quote Originally Posted by slapmaxwell1 View Post
    Consider L(y)=e^{2x}

    the Wronskian is W = \begin{vmatrix}<br />
e^{x} &  e^{2x}&e^{3x} \\ <br />
 e^{x}& 2e^{2x} & 3e^{2x}\\ <br />
 e^{x}&4e^{2x}  & 9e^{3x}<br />
\end{vmatrix}=2e^{6x}.

    ok so now I am supposed to write an annihilator for L(y)=0,
    then write the complementary solution y_{c},
    write an annihilator for L(y)=e^{2x}
    rewrite L(y)=e^{2x} in its expanded (y,y',...)
    notation.

    so the annihilator is (D-1)(D-2)(D-3)
    y_{c}=C_{1}e^{x}+C_{2}e^{2x}+C_{3}e^{3x}

    the last two parts of the problem im not sure how to answer, and to be honest im not sure if the work i have done up to this point is correct. any help would be appreciated.
    You are correct so far
    L(y)=(D-1)(D-2)(D-3)y Now we have

    (D-1)(D-2)(D-3)y=e^{2x} So we need to act on both sides by (D-2) this gives

    (D-2)(D-1)(D-2)(D-3)y=(D-2)e^{2x} \iff (D-1)(D-2)^2(D-3)y=0

    Now to write in expanded notation multiply out the operator to get

    (D-1)(D-2)(D-3)y=e^{2x} \iff (D^2-3D+2)(D-3)y=e^{2x}

    (D^3-6D^2+11D-6)y=e^{2x} \iff D^3y-6D^2y+11Dy-6y=e^{2x}

    y'''-6y''+11y'-6y=e^{2x}
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