Q 3.7-7E Use the fourth-order Runge鈥揔ut... [FREE SOLUTION]

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Fundamentals Of Differential Equations And Boundary Value Problems

R. Kent Nagle, Edward B. Saff, Arthur David Snider

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 616 pages

ISBN: 9780321977069

Chapter 3: Q 3.7-7E (page 90)

Use the fourth-order Runge鈥揔utta subroutine with h= 0.25 to approximate the solution to the initial value problem $y' = 2 y - 6, y (0) = 1$ , at x= 1. (Thus, input N= 4.) Compare this approximation to the actual solution $y = 3 - 2 e^{2 x}$ evaluated at x= 1.

Short Answer

Expert verified

$蠒 (1) = - 11.7679$

Step by step solution

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

Since $f (x, y) = 2 y - 6$ and x = 0, y = 1 and h = 0.25

$k_{1} = h f (x, y) = 0 . 25 (2 (1) - 6) = - 1 k_{2} = hf (x + \frac{h}{2}, y + \frac{k_{1}}{2}) = - 1.25 k_{3} = hf (x + \frac{h}{2}, y + \frac{k_{2}}{2}) = - 1.3125 k_{4} = hf (x + h, y + k_{3}) = - 1.65625$

Find the values of x and y

$x = 0 + 0.25 = 0.25 y = 1 + \frac{1}{6} (k_{1} + 2 k_{2} + 2 k_{3} + k_{4}) = 1 + \frac{1}{6} (- 1 - 2 (1 . 25) - 2 (1 . 3125) - 1 . 65625) = - 0.29688$

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

$k_{1} = h f (x, y) = 0 . 25 (2 (- 0 . 29688) - 6) = - 1 . 62844 k_{2} = hf (x + \frac{h}{2}, y + \frac{k_{1}}{2}) = - 2.06055 k_{3} = hf (x + \frac{h}{2}, y + \frac{k_{2}}{2}) = - 2.16358 k_{4} = hf (x + h, y + k_{3}) = - 2.73022$

Now

$x = 0.25 + 0.25 = 0.5 y = 1 + \frac{1}{6} (k_{1} + 2 k_{2} + 2 k_{3} + k_{4}) = 1 + \frac{1}{6} (- 1 . 64844 - 2 (2 . 06055 - 2 (2 . 16358 - 2 . 73022) = - 2.43470$

Repeat the procedure for two times

$k_{1} = h f (x, y) = 0 . 25 (2 (- 2 . 43470) - 6) = - 2 . 71735 k_{2} = hf (x + \frac{h}{2}, y + \frac{k_{1}}{2}) = - 3.39670 k_{3} = hf (x + \frac{h}{2}, y + \frac{k_{2}}{2}) = - 3.56652 k_{4} = hf (x + h, y + k_{3}) = - 4.50060 x = 0.5 + 0.25 = 0.75 y = 1 + \frac{1}{6} (k_{1} + 2 k_{2} + 2 k_{3} + k_{4}) = 1 + \frac{1}{6} (- 2 . 71735 - 2 (3 . 39670) - 2 (3 . 56652) - 4 . 50060) = - 5.95876$

And

$k_{1} = h f (x, y) = 0 . 25 (2 (- 5 . 95876) - 6) = - 4 . 47938 k_{2} = hf (x + \frac{h}{2}, y + \frac{k_{1}}{2}) = - 5.59922 k_{3} = hf (x + \frac{h}{2}, y + \frac{k_{2}}{2}) = - 5.87918 k_{4} = hf (x + h, y + k_{3}) = - 7.41895 x = 0.75 + 0.25 = 1 y = 1 + \frac{1}{6} (k_{1} + 2 k_{2} + 2 k_{3} + k_{4}) = - 5.95876 + \frac{1}{6} (- 4 . 47938 - 2 (5 . 59922) - 2 (5 . 87918) - 7 . 41895) = - 11.7679$

Since at x = 1

$蠒 (x) = 3 - 2 e^{2 x} 蠒 (1) = - 11.7781$

Then $|y (1) - 蠒 (1)| = 0.0120$

Hence the solution is $蠒 (1) = - 11.7679$

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91影视

Use the fourth-order Runge鈥揔utta subroutine with h= 0.25 to approximate the solution to the initial value problem $y' = 2 y - 6, y (0) = 1$ , at x= 1. (Thus, input N= 4.) Compare this approximation to the actual solution $y = 3 - 2 e^{2 x}$ evaluated at x= 1.

Short Answer

Step by step solution

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

Find the values of x and y

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

Repeat the procedure for two times

One App. One Place for Learning.

Most popular questions from this chapter

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Discrete Mathematics

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91影视

Use the fourth-order Runge鈥揔utta subroutine with h= 0.25 to approximate the solution to the initial value problem y'=2y-6,y(0)=1, at x= 1. (Thus, input N= 4.) Compare this approximation to the actual solution y=3-2e2xevaluated at x= 1.

Short Answer

Step by step solution

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

Find the values of x and y

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

Repeat the procedure for two times

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Probability and Statistics

Geometry

Pure Maths

Decision Maths

Calculus

Study anywhere. Anytime. Across all devices.

Use the fourth-order Runge鈥揔utta subroutine with h= 0.25 to approximate the solution to the initial value problem $y' = 2 y - 6, y (0) = 1$ , at x= 1. (Thus, input N= 4.) Compare this approximation to the actual solution $y = 3 - 2 e^{2 x}$ evaluated at x= 1.