Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 10 - Differential Equations - 10.3 Euler's Method - 10.3 Exercises - Page 550: 11


The actual result: $4.016$ The approximation: $4.023$ Difference: $4.023-4.016=0.007$

Work Step by Step

We are given $\frac{dy}{dx}=x^{3}$ Using Euler's method $g(x,y)=x^{3}$ Since $x=0, y=4$ $g(x_{0},y_{0})=0$ and $y_{1}=y_{0}+g(x_{0},y_{0})h=4+0\times0.1=4$ Now $x_{1}=0.1, y_{1}=4$ and $g(x_{1},y_{1})=0.001$ Then $y_{2}=4+(0.001)\times0.1=4.0001$ $y_{3}=4.0001+(0.008)\times0.1=4.0009$ $y_{4}=4.0009+(0.027)\times0.1=4.0036$ $y_{5}=4.0036+(0.064)\times0.1=4.01$ $y_{6}=4.01+(0.125)\times0.1=4.023$ $y(0.5)=4.023$ $\frac{dy}{dx}=x^{3}$ $\int dy=\int x^{3}dx$ $y=\frac{x^{4}}{4}+C$ With $x=0, y=4$ $C=4- \frac{0^{4}}{4}=4$ $y=\frac{x^{4}}{4}+4$ $y(0.5)=\frac{(0.5)^{4}}{4}+4=4.016$ Difference between the result and approximation: $4.023-4.016=0.007$
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