Answer
$y(x)=c_1x^6+c_2x^{-3}$
Work Step by Step
Given: $x^2y''-2xy'-18y=0$
In this case the substitution $y(x) = x^r$ yields the indicial equation
$r(r-1)-2r-18=0$
Factor and solve the equation
$r^2-3r-18=0$
It follows that two linearly independent solutions to the given differential equation are
$y_1(x)=x^6\\
y_2(x)=x^{-3}$
so that the general solution is
$y(x)=c_1x^6+c_2x^{-3}$
