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: 19


$\dfrac{1}{3} \ln |y^3-2|= x^3+c$

Work Step by Step

Given: $y^2 \dfrac{dy}{dx}=3x^2y^3-6x^2$ Re-arrange the given equation and integrate as follows:. Then $ \int 3x^2 dx =\int \dfrac{y^2}{(y^3-2)}dy\implies \int 3x^2 dx =(\dfrac{1}{3}) \int \dfrac{3y^2}{(y^3-2)}dy$ ....(1) As we know that $\int x^n dx=\dfrac{x^{n+1}}{n+1}+c$ Thus, $x^3+c =(\dfrac{1}{3}) \int \dfrac{3y^2}{(y^3-2)}dy$ ...(2) Let us take the help of $u$ substitution. plug $u=y^3 \implies du =3y^2 dy$ Equation (2) becomes: $\dfrac{1}{3} \int \dfrac{du}{u-2}=x^3+c \implies \dfrac{1}{3} \ln |u-2|= x^3+c$ Hence, $\dfrac{1}{3} \ln |y^3-2|= x^3+c$
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