Answer
See below
Work Step by Step
Given: $x^2y''-3xy'-12y=x^{-3}$
In this case the substitution $y(x) = x^r$ yields the indicial equation
$r(r-1)-3r-12=0$
Factor and solve the equation
$r^2-4r-12=0\\
(r+2)(r-6)=0$
It follows that two linearly independent solutions to the given differential equation are
$y_1(x)=x^{6}\\
y_2(x)=x^{-2}$
so that the particular solution is
$y(x)=u_1x^{6}+u_2x^{-2}$
We first compute the appropriate derivatives
$y'=u_1'(x)x^{6}+u_2'(x)x^{-2}\\
y''=u_1'(x)(x^{6})'+u_2'(x)(x^{-2})'$
Substituting these expressions into the given equation, we find that $u$ must satisfy
$u_1'=\frac{1}{8}(\frac{x^4+5x^2}{x^5})\\
u_1=\frac{1}{8}\ln x-\frac{5}{16x^2}$
and $u_1'=-\frac{1}{8}(x^7+5x^5)\\
u_1=-\frac{1}{64}x^8-\frac{5}{48}x^6$
Hence, the general solution is:
$y(x)=\frac{1}{8}x^6\ln x-\frac{1}{64}x^6-\frac{5}{12}x^4$
