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 531: 13


$y_1=4.2, y_2=6.216,y_3=9.6969$ and $y=3 e^{x^2+2x}$ and $y(0.6)=3 e^{(0.6)^2+2(0.6)} \approx 14.2765 $

Work Step by Step

The formula to calculate the approximations is defined as: $y_{n+1}=y_n+f(x_n,y_n) dx$ Consider $y(1)=0$ This gives us: $x_0=0,x_1=0+0.2=0.2, x_2=0.2+0.2=0.4$ Also, $y_1=4.2, y_2=6.216,y_3=9.6969$ Now, $\int \dfrac{dy}{y}=\int 2(x+1) dx$ or, $\ln |y|=x^2+2x+c$ when $y(1)=0$, then we have $c=\ln 3$ Thus, $\ln |y|=x^2+2x+c \implies \ln |y|=x^2+2x+\ln 3$ Here, $y=e^{x^2+2x+\ln 3}$ so, $y=3 e^{x^2+2x}$ Thus, $y(0.6)=3 e^{[(0.6)^2+2(0.6)]} \approx 14.2765 $
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