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 9.1 - Solutions, Slope Fields, and Euler's Method - Exercises 9.1 - Page 532: 36



Work Step by Step

Here, we have $\int \dfrac{dy}{y^2}=\int (2x-2) dx$ and $\dfrac{-1}{y}=x^2-2x+c$ Further, apply the initial conditions. $-\dfrac{1}{2}=2^2-2(2)+c=4-4+c$ or, $c=2$ Now, we have $\dfrac{-1}{y}=x^2-2x+c \implies \dfrac{-1}{y}=x^2-2x+2$ Thus, $y=-\dfrac{1}{x^2-2x+2}$ and $y(3)=-\dfrac{1}{3^2-2(3)+2} =-\dfrac{1}{9-6+2} = -0.2$
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