Answer
$y(x)=c_1\cos (3\ln x)+c_2\sin (3\ln x)-\cos (3x \ln x)\cos (3\ln x)+\sin (3x\ln x)\sin (3\ln x)$
Work Step by Step
Given $x^2y''+xy'+9y=9\ln x$
In this case the substitution $y(x) = x^r$ yields the indicial equation
$$r(r-1)+r+9=0\\
r^2+9=0$$
It follows that two linearly independent solutions to the given differential equation are
$u_1(x)=\cos (3 \ln x)\\
u_2(x)=\sin (3\ln x)$
so that the general solution is
$y(x)=u_1(x)\cos (3\ln x)+u_2(x)\sin (3\ln x)$
where $u_1$ and $u_2$ are determined from
$u_1'(x)\cos (3\ln x)+u_2'(x)\sin (3\ln x)=0\\
u_1'(x)(\cos (3\ln x))'+u'_2(x)(\sin (3\ln x))'=9x^{-2}\ln x$
Hence,
$u_1'=3x^{-1}\cos (3\ln x)\ln x\\
u_2'=-3x^{-1}\sin (3\ln x)\ln x$
which upon integration gives
$u_1(x)=\ln x\cos (3\ln x)-\frac{1}{3}\sin(3\ln x)\\
u_2(x)=\ln x\sin (3\ln x)+\frac{1}{3}\cos (3\ln x)$
Thus, the given equation has general solution
$y(x)=c_1\cos (3\ln x)+c_2\sin (3\ln x)-\cos (3x \ln x)\cos (3\ln x)+\sin (3x\ln x)\sin (3\ln x)$
