Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 17 - Section 17.1 - Second-Order Linear Equations - 17.1 Exercise - Page 1160: 9


y = $C_{1}e^{2x}cos(3x)$ + $C_{2}e^{2x}sin(3x)$

Work Step by Step

Question: y"-4y'+13y=0 We know: y = $e^{rx}$ y' = $re^{rx}$ y" = $r^{2}e^{rx}$ This results in $(r^{2}-4r+13)e^{rx}$=0 The corresponding characteristic equation is: $r^{2}-4r+13=0$ Solving the characteristic equation with the abc-formula and complex numbers gives: $\frac{-b\frac{+}{-}\sqrt {b^{2}-4*a*c}}{2*a}$ $\frac{4\frac{+}{-}\sqrt {(-4)^{2}-4*1*13}}{2*1}$ $r=2\frac{+}{-}3i$ Adding the constants gives the following general solution: y = $C_{1}e^{2x}cos(3x)$ + $C_{2}e^{2x}sin(3x)$
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