Answer
True
Work Step by Step
For $x=r^2 $
$r(r-1) -2r-18=0 \implies (r-6)(r+3)=0$
So, $r_1=6;r_2=-3$
Thus, $x^6$ and $x^{-3}$ are linearly independent solution.
Therefore, the given statement is $\bf{True}$.
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