Thomas' Calculus 13th Edition

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

Chapter 4: Applications of Derivatives - Practice Exercises - Page 245: 79

Answer

$x-\dfrac{1}{x}-1$

Work Step by Step

As we know that $\int x^{a} dx=\dfrac{x^{a+1}}{a+1}+C$ Now, we have $\dfrac{dy}{dx}=\dfrac{x^2+1}{x^2}$ and $y(1)=-1$ This implies that $y=\int \dfrac{x^2+1}{x^2} dx=\int (1+\dfrac{1}{x^2}) dx$ $ x+[\dfrac{x^{(-2+1)}}{(-2+1)}]+C=x-(\dfrac{1}{x})+C$ Apply the initial condition, we get $y(1)=-1$ Then $C=-1$ Hence, $y=x-(\dfrac{1}{x})-1$
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