Answer
True
Work Step by Step
Since, $D^2y-y=0 $
Plug $y=e^{rx} $
So, $(r^2-1)e^{rx}=0$
This implies that $r^2-1=0 \implies r=\pm 1$ and $y=c_1e^x+c_2e^{-x}$
So, basis of $ker(L)={e^x, e^{-x}}$
Therefore, the given statement is 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 - 8.1 General Theory for Linear Differential Equations - True-False Review - Page 503: h
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