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

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

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

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

ok so now I am supposed to write an annihilator for L(y)=0,
then write the complementary solution $y_{c}$,
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.

2. Originally Posted by slapmaxwell1
Consider $L(y)=e^{2x}$

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

ok so now I am supposed to write an annihilator for L(y)=0,
then write the complementary solution $y_{c}$,
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

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

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

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