University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 16 - Section 16.1 - Solutions, Slope Fields, and Euler's Method - Exercises - Page 16-8: 13

Answer

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

Work Step by Step

Apply the formula to find the approximations as follows: $y_{n+1}=y_n+f(x_n,y_n) dx$ ....(1) We are given $y(1)=0$ This implies $x_0=0,x_1=0+0.2=0.2, x_2=0.2+0.2=0.4$ Likewise, by using equation (1), we have $y_1=4.2, y_2=6.216,y_3=9.6969$ In order to determine the exact solution, isolate the x and y terms on one side and integrate both sides. $\int \dfrac{dy}{y}=\int 2(x+1) dx$ ...(2) In order to determine the differential equation, we will have to take the derivative of the differential equation. Equation (2) gives: $\ln |y|=x^2+2x+c$ ...(3) Apply the initial conditions to calculate the value of $c$ $\ln |3|=(0)^2+2(0)+c \implies c=\ln 3$ Now, equation (3) becomes: $\ln |y|=x^2+2x+\ln 3$ This implies that $y=e^{x^2+2x+\ln 3} \implies y=3 e^{x^2+2x}$ Now, $y(0.6)=3 e^{(0.6)^2+2(0.6)} \approx 14.2765 $ The accuracy of the approximation can be found as: $14.2765-96969=4.5796$ Hence, our answers are: $y_1=4.2, y_2=6.216,y_3=9.6969$ and $y=3 e^{x^2+2x}$ and the accuracy is: $4.5796$
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