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: 35



Work Step by Step

Here, we have $\int dy=\int 2xe^{x^2} dx$ Now, $y=e^{x^2}+c$ Further, apply the initial conditions. $2=e^{(0^2)}+c$ or, $c=1$ Thus, $y=e^{x^2}+c \implies y=e^{x^2}+1$ Therefore, $y(1)=e^{(1)^2}+1 =e+1 \approx 3.71728$
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