Answer
$y(x)=c_1\cos (4\ln x)+c_2\sin (4\ln x)$
Work Step by Step
Given $x^2y''+xy'+16y=0$
In this case the substitution $y(x) = x^r$ yields the indicial equation
$$r(r-1)+r+16=0\\
r^2+16=0$$
It follows that two linearly independent solutions to the given differential equation are
$y_1(x)=\cos (4\ln x)\\
y_2(x)=\sin (4\ln x)$
so that the general solution is
$y(x)=c_1\cos (4\ln x)+c_2\sin (4\ln x)$
