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Q5.3-21E

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Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 260

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

Fluid Ejection. In the design of a sewage treatment plant, the following equation arises: $60 - H = (77 . 7) H'' + (19 . 42) (H')^{2}; H (0) = H' (0) = 0$ where H is the level of the fluid in an ejection chamber, and t is the time in seconds. Use the vectorized Runge–Kutta algorithm with h = 0.5 to approximate $H (t)$ over the interval [0, 5].

The result can get by the Runge-Kutta method.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Transform the equation

Here the equation is $60 - H = (77 . 7) H'' + (19 . 42) (H')^{2}; H (0) = H' (0) = 0 .$

The system can be written as:

$x_{1} = H (t) x_{2} = H' = x'_{1} H'' = x'_{2}$

The transform equation:

$x'_{1} = x_{2} x'_{2} = \frac{[60 - x_{1} - (19 . 42) {x^{2}}_{2}]}{77.7}$

The initial conditions are:

$x_{1} (0) = H (0) = 0 x_{2} (0) = H' (0) = 0$

Apply the Runge-Kutta method.

Here h=0.5, N=10 steps, $x_{1, 0} = 0, x_{2, 0} = 0$ then;

$x_{1, 0} = 0, x_{2, 0} = 0 k_{1, 1} = h x_{2, 0} = 0 . 5 (0) = 0 k_{2, 1} = \frac{h [60 - x_{1, 0} - (19 . 42) {x^{2}}_{2, 0}]}{77.7} = 0.38610 k_{1, 2} = 0.09653 k_{2, 2} = 0.38144 k_{1, 3} = 0.09536 k_{2, 3} = 0.38124 k_{1, 4} = 0.19062 k_{2, 4} = 0.36732$

Apply the recursive formula

$t_{1} = t_{0} + h 0 + 0.5 = 0.5 x_{1} (0 . 5) = x_{1, 1} = 0.09573 x_{2} (0 . 5) = x_{2, 1} = 0.37980$

Apply the same procedure for n=1, 2,…., 9.

N	$t_{n}$	$x_{1, n} = H (t_{n})$
0	0	0
1	0.5	0.09573
2	1.0	0.37389
3	1.5	0.81045
4	2.0	1.37361
5	2.5	2.03111
6	3.0	2.7549
7	3.5	3.52322
8	4.0	4.319
9	4.5	5.13307
10	5.0	5.95554

This is the required result.

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