Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 9: First-Order Differential Equations - Practice Exercises - Page 557: 10


$y=e^{-x}(\sin x-\cos x)+ce^{-2x}$

Work Step by Step

Here, we have $y'+2y=2e^{-x} \sin x$ The integrating factor is: $I=e^{\int 2 dx}=e^{2x}$ Now, $e^{2x}[y'+2y]=2 (e^{2x}) (e^{-x}) (\sin x)$ This implies that $\int [e^{2x}y]' =\int 2e^{x}\sin x dx$ Hence, $y=e^{-x}(\sin x-\cos x)+ce^{-2x}$
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