Answer
$y(x)=c_1x^{-2}+c_2x^{-3}-2e^{2x}(x-1)x^{-3}$
Work Step by Step
Given $x^2y''+6xy'+6y=4e^{2x}$
In this case the substitution $y(x) = x^r$ yields the indicial equation
$$r(r-1)+6r+6=0\\
r^2+5r+6=0$$
It follows that two linearly independent solutions to the given differential equation are
$u_1(x)=x^{-2}\\
u_2(x)=x^{-3}$
so that the general solution is
$y(x)=u_1(x)x^{-2}+u_2(x)x^{-3}$
where $u_1$ and $u_2$ are determined from
$u_1'(x)x^{-2}+u_2'(x)x^{-3}=0\\
u_1'(x)(x^{-2})'+u'_2(x)(x^{-3})'=4e^{2x}$
Hence,
$u_1'=4xe^{2x}\\
u_2'=-4x^2e^{2x}$
which upon integration gives
$u_1(x)=2xe^{2x}-e^{2x}\\
u_2(x)=-2x^2e^{2x}+2xe^{2x}-e^{2x}$
Thus, the given equation has general solution
$y(x)=c_1x^{-2}+c_2x^{-3}-2e^{2x}(x-1)x^{-3}$
