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: 9


$y=x(\ln x)^2+xc$

Work Step by Step

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