University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.2 - Exponential Change and Separable Differential Equations - Exercises - Page 409: 20


$\dfrac{x^{2}}{2}-2x -\ln |y+3|= c$

Work Step by Step

Given: $\dfrac{dy}{dx}=xy+3x-2y-6$ Re-arrange the given equation and integrate as follows:. Then $ \int \dfrac{dy}{dx}=\int (x-2)(y+3) \implies \int (x-2) dx = \int \dfrac{dy}{y+3}$ ....(1) As we know that $\int x^n dx=\dfrac{x^{n+1}}{n+1}+c$ Thus, $\dfrac{x^{1+1}}{1+1}-2x =\ln |y+3| +c$ $\dfrac{x^{2}}{2}-2x =\ln |y+3| +c$ Hence, $\dfrac{x^{2}}{2}-2x -\ln |y+3|= c$
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