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 fourth-order Runge–Kutta subroutine with h = 0.25 to approximate the solution to the initial value problem $y' = x + 1 - y, y (0) = 1$ , at x = 1. Compare this approximation with the one obtained in Problem 5 using the Taylor method of order 4.

$ϕ (1) = 1.3679$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Find the values of ki, i = 1, 2, 3, 4

Since $f (x, y) = x + 1 - y$ and x = 0, y = 1, and h = 0.25

role="math" localid="1664324055813" $k_{1} = h f (x, y) = 0 . 25 (0 + 1 - 1) = 0 k_{2} = hf (x + \frac{h}{2}, y + \frac{k_{1}}{2}) = 0.03125 k_{3} = hf (x + \frac{h}{2}, y + \frac{k_{2}}{2}) = 0.0273438 k_{4} = hf (x + h, y + k_{3}) = 0.0556641$

Step 2: Find the values of x and y

$x = 0 + 0.25 = 0.25 y = 1 + \frac{1}{6} (k_{1} + 2 k_{2} + 2 k_{3} + k_{4}) = 1 + \frac{1}{6} (0 - 2 (0 . 031125) - 2 (0 . 0273438) - 0 . 0556641) = 1.02881$

Step 3: Use values of x and y for finding values of ki, i = 1, 2, 3, 4

$k_{1} = h f (x, y) = 0 . 25 (0 . 25 + 1 - 1 . 02881) = 0 . 0552975 k_{2} = hf (x + \frac{h}{2}, y + \frac{k_{1}}{2}) = 0.0796353 k_{3} = hf (x + \frac{h}{2}, y + \frac{k_{2}}{2}) = 0.0765931 k_{4} = hf (x + h, y + k_{3}) = 0.0986392$

Now

$x = 0.25 + 0.25 = 0.5 y = 1.02881 + \frac{1}{6} (0 . 0552975 - 2 (0 . 0796353) - 2 (0 . 0765931) - 0 . 0986492) = 1.10654$

Step 4: Repeat the procedure for two times

$k_{1} = h f (x, y) = 0 . 25 (0 . 5 + 1 - 1 . 10654) = 0 . 098365 k_{2} = hf (x + \frac{h}{2}, y + \frac{k_{1}}{2}) = 0.117319 k_{3} = hf (x + \frac{h}{2}, y + \frac{k_{2}}{2}) = 0.11495 k_{4} = hf (x + h, y + k_{3}) = 0.132127 x = 0.5 + 0.25 = 0.75 y = 1.10654 + \frac{1}{6} (k_{1} + 2 k_{2} + 2 k_{3} + k_{4}) = 1.10654 + \frac{1}{6} (0 . 098365 - 2 (0 . 117319) - 2 (0 . 11495) - 0 . 132127) = 1.22238$

And

$k_{1} = h f (x, y) = 0 . 25 (0 . 75 + 1 - 1 . 22238) = 0 . 131905 k_{2} = hf (x + \frac{h}{2}, y + \frac{k_{1}}{2}) = 0.146667 k_{3} = hf (x + \frac{h}{2}, y + \frac{k_{2}}{2}) = 0.144822 k_{4} = hf (x + h, y + k_{3}) = 0.1582 x = 0.75 + 0.25 = 1 y = 1.22238 + \frac{1}{6} (k_{1} + 2 k_{2} + 2 k_{3} + k_{4}) = 0.52761 + \frac{1}{6} (0 . 131905 - 2 (0 . 146667) - 2 (0 . 144822) - 0 . 1582) = 1.36789$

Therefore $ϕ (1) = 1.3679$

Hence the solution is $ϕ (1) = 1.3679$

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Use the fourth-order Runge–Kutta subroutine with h = 0.25 to approximate the solution to the initial value problem $y' = x + 1 - y, y (0) = 1$ , at x = 1. Compare this approximation with the one obtained in Problem 5 using the Taylor method of order 4.

Step by Step Solution

Step 1: Find the values of ki, i = 1, 2, 3, 4

Step 2: Find the values of x and y

Step 3: Use values of x and y for finding values of ki, i = 1, 2, 3, 4

Step 4: Repeat the procedure for two times

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Use the fourth-order Runge–Kutta subroutine with h = 0.25 to approximate the solution to the initial value problem y'=x+1-y,y(0)=1, at x = 1. Compare this approximation with the one obtained in Problem 5 using the Taylor method of order 4.

Step by Step Solution

Step 1: Find the values of ki, i = 1, 2, 3, 4

Step 2: Find the values of x and y

Step 3: Use values of x and y for finding values of ki, i = 1, 2, 3, 4

Step 4: Repeat the procedure for two times

Most popular questions for Math Textbooks

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Pure Maths

Use the fourth-order Runge–Kutta subroutine with h = 0.25 to approximate the solution to the initial value problem $y' = x + 1 - y, y (0) = 1$ , at x = 1. Compare this approximation with the one obtained in Problem 5 using the Taylor method of order 4.