Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.4 - Exponential Change and Separable Differential Equations - Exercises 7.4 - Page 400: 19


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

Work Step by Step

Rewrite the given equation and then integrate. We have $ \int 3x^2 dx =\int \dfrac{y^2}{(y^3-2)}dy$ This can also be written as: $\int 3x^2 dx =(\dfrac{1}{3}) \int \dfrac{3y^2}{(y^3-2)}dy$ Now, $x^3+c =\dfrac{1}{3} \int \dfrac{3y^2}{(y^3-2)}dy$ Suppose $a=y^3 \implies da =3y^2 dy$ $x^3+c =\dfrac{1}{3} \int \dfrac{3y^2}{(y^3-2)}dy \implies \dfrac{1}{3} \int \dfrac{da}{a-2}=x^3+c \implies \dfrac{1}{3} \ln |a-2|= x^3+c$ Hence, $\dfrac{1}{3} \ln |y^3-2|= x^3+c$
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