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: 9



Work Step by Step

Here, we have $y'-\dfrac{1}{2}y=\dfrac{x}{2}e^{(x/2)}$ The integrating factor is: $I=e^{\int (-1/2)dx}=e^{-x/2}$ Now, $e^{-x/2}[y'-\dfrac{y}{2}]=(e^{-x/2})[(\dfrac{x}{2})(e^{(x/2)})]$ This implies that $\int [e^{-(x/2)}y]' =\int [(\dfrac{x}{2})(e^{(x/2)})] (e^{-x/2}) dx$ Hence, $y=e^{(x/2)}[\dfrac{x^2}{4}+c]$
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