Answer
True
Work Step by Step
For $y=x^r$
$r(r-1)+6r+6=0 \implies (r+2)(r+3)=0$
So, $r_1=-2;r_2=-3$
Thus, $x^{-2}$ and $x^{-3}$ are linearly independent solution.
Therefore, the given statement is $\bf{True}$.
Differential Equations and Linear Algebra (4th Edition)
by Goode, Stephen W.; Annin, Scott A.
Published by
Pearson
ISBN 10:
0-32196-467-5
ISBN 13:
978-0-32196-467-0
Chapter 8 - Linear Differential Equations of Order n - 8.8 A Differential Equation with Nonconstant Coefficients - True-False Review - Page 567: d
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