Answer
See below
Work Step by Step
Given: $y''-6y'+9y=4e^{3x}\ln x$
In this case, the auxiliary polynomial of the associated homogeneous equation is
$P(r) = r^2 − 6r + 9r =0\\
\rightarrow (r-3)^2=0$
so that three linearly independent solutions are:
$y_1(x)=e^{3x}\\
y_2(x)=xe^{3x}$
A particular solution is:
$y_p(x)=e^{3x}u_1(x)+xe^{3x}u_2(x)$
where $u_1,u_2$ satisfy the form:
$e^{3x}u_1'(x)+xe^{3x}u_2'(x)=0\\
(e^{3x})^2u_1'(x)+(xe^{3x})^2u_2'(x)=0$
Consequently $u_1'(x)=-xu_2'(x)\\
u_2'(x)=4 \ln x$
By integrating, we obtain
$u_2(x)=4\int \ln x dx=4(x\cos x-x)+C_1$
$u_1'(x)=-xu_2'(x)=-4x\ln x \rightarrow u_1=\int -4x\ln x dx=-2x^2\ln x+x^2+C_2$
where $C_1,C_2$ are integration constants.
The general solution to the given differential equation is therefore:
$y(x)=(-2x^2 \ln x+x^2+C_2)e^{3x}+xe^{3x}(4x \ln x -x+C_1)\\
=e^{3x}(2x^2\ln x+C_1x+C_2)$
