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Q 3.6-8E

Expert-verified

Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 130

Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

Use the improved Euler’s method subroutine with step size h = 0.2 to approximate the solution to the initial value problem $y' = \frac{1}{x} (y^{2} + y), y (1) = 1$ at the points x = 1.2, 1.4, 1.6, and 1.8. (Thus, input N = 4.) Compare these approximations with those obtained using Euler’s method (see Exercises 1.4, Problem 6, page 28).

x_n	y_n
1.2	1.48
1.4	2.24788
1.6	3.6518
1.8	6.88733

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Find the equation of approximation value

Here, $y' = \frac{1}{x} (y^{2} + y), y (1) = 0$ for $1 ⩽ x ⩽ 1.8$

For h=0.2, x=1, y=1, N=4

$F = f (x, y) = \frac{1}{x} (y^{2} + y) G = f (x + h, y + hF) = \frac{1}{x + 0.2} ({(y + \frac{0.2}{x} (y^{2} + y))}^{2} + y + \frac{0.2}{x} (y^{2} + y))$

Step 2: Solve for x1 and y1

Apply initial points $x_{o} = 1, y_{o} = 1, h = 0.2$

$F (1, 1) = 2 G (1, 1) = 2 . 8$

$x_{1} = 1 + 0.2 = 1.2 y_{1} = 1 + \frac{0.2}{2} (2 + 2 . 8) = 1.48$

Step 3: Evaluate the value of x2 and y2

$F (1 . 2, 1 . 48) = 3 . 05867 G (1 . 2, 1 . 48) = 4 . 61934$

$x_{2} = 1.2 + 0.2 = 1.4 y_{2} = 1.48 + 0 . 1 (3 . 05867 + 4 . 61934) = 2.24788$

Step 4: Determine the value of x3 and y3

$F (1 . 4, 2 . 24788) = 5 . 21489 G (1 . 4, 2 . 2488) = 8 . 82538$

$x_{3} = 1.4 + 0.2 = 1.6 y_{3} = 2.24788 + 0 . 1 (5 . 21489 + 8 . 82538) = 3.6518$

Step 5: Determine the value of x4 and y4

$F (1 . 6, 3 . 6518) = 10 . 6172 G (1 . 6, 3 . 6518) = 21 . 7381$

$x_{4} = 1.6 + 0.2 = 1.8 y_{4} = 3.6518 + 0 . 1 (10 . 6172 + 21 . 7381) = 6.88733$

Hence the solution is

x_n	y_n
1.2	1.48
1.4	2.24788
1.6	3.6518
1.8	6.88733

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