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Q5.3-27E

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Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 261

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

Generalized Blasius Equation. H. Blasius, in his study of the laminar flow of a fluid, encountered an equation of the form $y''' + yy'' = (y')^{2} - 1$ . Use the Runge–Kutta algorithm for systems with h = 0.1 to approximate the solution that satisfies the initial conditions $y (0) = 0, y' (0) = 0, y'' (0) = 1 . 32824$ . Sketch this solution on the interval [0, 2].

The result can get by the Runge-Kutta method, and the result is y(2)=1.6001.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Transform the equation

Here the equation is $y''' + yy'' = (y')^{2} - 1$ .

The system can be written as:

$x_{1} = y x_{2} = y' = x'_{1} x_{3} = y'' = x'_{2}$

The transform equation is:

$x'_{1} = x_{2} x'_{2} = x_{3} x'_{3} = - x_{1} x_{3} + {x^{2}}_{2} - 1$

The initial conditions are:

$x_{1} (0) = y (0) = 0 x_{2} (0) = y' (0) = 0 x_{3} (0) = y'' (0) = 1 . 32824$

Apply the Runge-Kutta method

For h=0.1

t	Y	T	Y
0	0	1.1	0.599
0.1	0.00647	1.2	0.69515
0.2	0.0252	1.3	0.79515
0.3	0.0553	1.4	0.89926
0.4	0.0957	1.5	1.0072
0.5	0.1456	1.6	1.1189
0.6	0.20407	1.7	1.234
0.7	0.27032	1.8	1.3526
0.8	0.34363	1.9	1.4747
0.9	0.4233	2	1.6001
1	0.50882

Graph

Therefore, the value of y(2)=1.6001.

Thus, this is the required result.

Most popular questions for Math Textbooks

A Problem of Current Interest. The motion of an ironbar attracted by the magnetic field produced by a parallel current wire and restrained by springs (see Figure 5.17) is governed by the equation\(\frac{{{{\bf{d}}^{\bf{2}}}{\bf{x}}}}{{{\bf{d}}{{\bf{t}}^{\bf{2}}}}}{\bf{ = - x + }}\frac{{\bf{1}}}{{{\bf{\lambda - x}}}}\) for \({\bf{ - }}{{\bf{x}}_{\bf{o}}}{\bf{ < x < \lambda }}\)where the constants \({{\bf{x}}_{\bf{o}}}\) and \({\bf{\lambda }}\) are, respectively, the distances from the bar to the wall and to the wire when thebar is at equilibrium (rest) with the current off.

Setting\({\bf{v = }}\frac{{{\bf{dx}}}}{{{\bf{dt}}}}\), convert the second-order equation to an equivalent first-order system.
Solve the related phase plane differential equation for the system in part (a) and thereby show that its solutions are given by\({\bf{v = \pm }}\sqrt {{\bf{C - }}{{\bf{x}}^{\bf{2}}}{\bf{ - 2ln(\lambda - x)}}} \), where C is a constant.
Show that if \({\bf{\lambda < 2}}\) there are no critical points in the xy-phase plane, whereas if \({\bf{\lambda > 2}}\) there are two critical points. For the latter case, determine these critical points.
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From your phase plane diagrams in part (d), describe the possible motions of the bar when \({\bf{\lambda = 1}}\) and when\({\bf{\lambda = 3}}\), under various initial conditions.

In Problems 11–14, solve the related phase plane differential equation for the given system. Then sketch by hand several representative trajectories (with their flow arrows).

$\frac{d x}{d t} = \frac{3}{y} \frac{d y}{d t} = \frac{2}{x}$

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$\frac{d x}{d t} = - 5 x + 2 y, \frac{d y}{d t} = x - 4 y$

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$\frac{dx}{dt} + y = t^{2}, - x + \frac{dy}{dt} = 1$

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