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Thread: the Wronskian and variation of parameters...

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

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

    the Wronskian is W = $\displaystyle \begin{vmatrix}
    e^{x} & e^{2x}&e^{3x} \\
    e^{x}& 2e^{2x} & 3e^{2x}\\
    e^{x}&4e^{2x} & 9e^{3x}
    \end{vmatrix}=2e^{6x}.$

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

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

    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 $\displaystyle L(y)=e^{2x}$

    the Wronskian is W = $\displaystyle \begin{vmatrix}
    e^{x} & e^{2x}&e^{3x} \\
    e^{x}& 2e^{2x} & 3e^{2x}\\
    e^{x}&4e^{2x} & 9e^{3x}
    \end{vmatrix}=2e^{6x}.$

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

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

    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
    $\displaystyle L(y)=(D-1)(D-2)(D-3)y$ Now we have

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

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

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

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

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

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