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 - Section 9.2 - First-Order Linear Equations - Exercises 9.2 - Page 536: 7



Work Step by Step

Re-write the given differential equation as: $y'-\dfrac{y}{2}=\dfrac{e^{x/2}}{2}$ The integrating factor is: $v(x)=e^{\int (-1/2) dx}=e^{-(x/2)}$ Now, we have $\int [(e^{-x/2})y)' dx=\int \dfrac{1}{2} dx$ or, $y=e^{x/2}(\dfrac{x}{2})+c$ Hence, the General solution is: $y=\dfrac{xe^{x/2}}{2}+ce^{x/2}$
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