Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 17 - Second-Order Differential Equations - Review - Exercises - Page 1193: 12

Answer

$y=2e^{3x}cos4x+\frac{-5}{4}e^{3x}sin4x$

Work Step by Step

$y''-6y'+25y=0$ which has characteristic equation $t^2-6t+25=0$ with solutions $t=3 \pm 4i$ General solution: $c_1e^{3x}cos(4x+c_2e^{3x}sin(4x)$ Plug in the values: $y(0)=2$ and $y'(0)=1$ giving $c_1=2$, $c_2=\frac{-5}{4}$ Hence, $y=2e^{3x}cos4x+\frac{-5}{4}e^{3x}sin4x$
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