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: 14


$y_1=0.5, y_2=0.5,y_3=0.625$, $y=\dfrac{1}{-x^2-x+1}$, and $y(0.5)=\dfrac{1}{-(0.5)^2-(0.5)+1} \approx 4$

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 $y(-1)=1$ This implies $x_0=-1,x_1=-1+0.5=-0.5, x_2=-0.5+0.5=0$ Likewise, by using equation (1), we have $y_1=0.5, y_2=0.5,y_3=0.625$ In order to determine the exact solution, isolate the x and y terms on one side and integrate both sides. $\int \dfrac{dy}{y^2}=\int (1+2x) dx$ ...(2) In order to determine the differential equation, we will have to take the derivative of the differential equation. Equation (2) gives: $y=\dfrac{1}{-x^2-x+c}$ ...(3) Apply the initial conditions, to calculate the value of $c$ $1=\dfrac{1}{-(-1)^2-(-1)+c} \implies c=1$ Now, equation (3) becomes: $y=\dfrac{1}{-x^2-x+1}$ Now, $y(0.5)=\dfrac{1}{-(0.5)^2-(0.5)+1} \approx 4$ The accuracy of the approximation can be found as: $4-0.625=3.375$ Hence, our answers are: $y_1=0.5, y_2=0.5,y_3=0.625$ and $y=\dfrac{1}{-x^2-x+1}$ and the accuracy is: $3.375$
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