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: 38



Work Step by Step

Here, $\int \dfrac{1}{1+y^2}dy=\int dx$ and $\arctan (y)=x+c$ Further, apply the initial conditions, to calculate the value of $c$ $\arctan (0)=0+c$ or, $c=0$ Now, the particular solution as follows: $\arctan (y)=x$ This can be re-arranged as: $y=\tan x$ and $y(1)=\tan x =\tan (1) \approx 1.5574$
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