Answer
$y(x)=c_1x^{-4}+c_2x^{-4}\ln x$
Work Step by Step
Given: $x^2y''+9xy'+16y=0$
In this case the substitution $y(x) = x^r$ yields the indicial equation
$r(r-1)+9r+16=0$
Factor and solve the equation
$r^2+8r+16=0$
It becomes
$(r+4)^2=0$
It follows that two linearly independent solutions to the given differential equation are
$y_1(x)=x^{-4}\\
y_2(x)=x^{-4}\ln x$
so that the general solution is
$y(x)=c_1x^{-4}+c_2x^{-4}\ln x$
