Answer
False
Work Step by Step
For $y=x^r$
$r(r-1)+9r+16=0 \implies (r+4)(r+4)=0$
So, $r_1=-4;r_2=-4$
Thus, $x^{-4}$ and $x^{-4}$ are linearly independent solution which does not have the form of $x^r$.
Therefore, the given statement is $\bf{False}$.