Elementary Differential Equations and Boundary Value Problems 9th Edition

Published by Wiley
ISBN 10: 0-47038-334-8
ISBN 13: 978-0-47038-334-6

Chapter 3 - Second Order Linear Equations - 3.3 Complex Roots of the Characteristic Equation - Problems - Page 163: 5

Answer

$2(cos(ln(2))-isin(ln(2)))$

Work Step by Step

$2^{1-i} = 2^1 2^{-i}$ Applying the identity $a^{ib} = cos(ln(a)b) + isin(ln(a)b)$ yields: $cos(-ln(2)) + isin(-ln(2))$ Cosine and sine are even and odd functions, respectively, so $cos(-ln(2)) + isin(-ln(2)) = cos(ln(2)) -isin(ln(2))$ In total: $2(cos(ln(2))-isin(ln(2)))$
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