Answer
See below
Work Step by Step
Given: $x^2y''-x(\cos x)y'+5e^{2x}y=0$
We can notice that $p=\cos x, q=5e^{2x}$
and $p(0)=-1, q(0)=5$
Obtain: $r(r-1)-r+5=0\\
r^2-2r+5=0\\
(r-1)^2+4=0\\
r-1=\pm 2i\\
\rightarrow r=1 \pm 2i$
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 11 - Series Solutions to Linear Differential Equations - 11.4 Series Solutions about a Regular Singular Point - Problems - Page 758: 9
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