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Q28E

Expert-verified

Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 250

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

In Problems 25 – 28, use the elimination method to find a general solution for the given system of three equations in the three unknown functions x(t), y(t), and z(t).
$x' = x + 2 y + z, y' = 6 x - y, z' = - x - 2 y - z$

The solutions for the given linear system are $x (t) = - \frac{1}{2} {Ae}^{- 4 t} + \frac{2}{3} {Be}^{3 t} + \frac{C}{6}$ ,

$y (t) = {Ae}^{- 4 t} + {Be}^{3 t} + C$ and $z (t) = \frac{1}{2} {Ae}^{- 4 t} - \frac{2}{3} {Be}^{3 t} - \frac{13}{6} C$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: General form

Elimination Procedure for 2 × 2 Systems:

To find a general solution for the system;

$L_{1} [x] + L_{2} [y] = f_{1}, L_{3} [x] + L_{4} [y] = f_{2},$

Where $L_{1}, L_{2}, L_{3},$ and $L_{4}$ are polynomials in $D = \frac{d}{dt}$

Make sure that the system is written in operator form.

Eliminate one of the variables, say, y, and solve the resulting equation for x(t). If the system is degenerating, stop! A separate analysis is required to determine whether or not there are solutions.

(Shortcut) If possible, use the system to derive an equation that involves y(t) but not its derivatives. [Otherwise, go to step (d).] Substitute the found expression for x(t) into this equation to get a formula for y(t). The expressions for x(t), and y(t) give the desired general solution.

Eliminate x from the system and solve for y(t). [Solving for y(t) gives more constants----in fact, twice as many as needed.]

Remove the extra constants by substituting the expressions for x(t) and y(t) into one or both of the equations in the system. Write the expressions for x(t) and y(t) in terms of the remaining constants.

Step 2: Evaluate the given equation

Given that,

$x' = x + 2 y + z$ … (1)

$y' = 6 x - y$ … (2)

$z' = - x - 2 y - z$ … (3)

Let us rewrite the given system of equations into operator form.

$(D - 1) [x] - 2 [y] - [z] = 0$ … (4)

$6 x - (D + 1) [y] = 0$ … (5)

$[x] + 2 [y] + (D + 1) [z] = 0$ … (6)

Multiply D+1 on equation (4). Then, add with equation (6).

$(D - 1) (D + 1) [x] - 2 (D + 1) [y] - (D + 1) [z] + [x] + 2 [y] + (D + 1) [z] = 0 (D^{2} - 1) [x] + [x] - 2 (D + 1) [y] + 2 [y] = 0 (D^{2} - 1 + 1) [x] + (- 2 D - 2 + 2) [y] = 0 D^{2} [x] - 2 D [y] = 0 D^{2} [x] - 2 D [y] = 0 \dots (7)$

Now multiply D²on equation (5) and 6 on equation (7). Then, subtract them.

$6 D^{2} [x] - 12 D [y] - 6 D^{2} [x] + D^{2} (D + 1) [y] = 0 (D^{3} + D^{2} - 12 D) [y] = 0 D (D^{2} + D - 12) [y] = 0 D (D + 4) (D - 3) [y] = 0$

Step 3: Substitution method

Since the auxiliary equation to the corresponding homogeneous equation is $r^{3} + r^{2} - 12 r = 0$ . The roots are $r = 0, r = - 4$ and $r = 3$ .

Then, the general solution of y is;

$y (t) = {Ae}^{- 4 t} + {Be}^{3 t} + C \dots (8)$

Now substitute equation (8) in equation (5).

$6 x - (D + 1) [y] = 0 6 x = (D + 1) [y] = (D + 1) [{Ae}^{- 4 t} + {Be}^{3 t} + C] = - 4 {Ae}^{- 4 t} + 3 {Be}^{3 t} + {Ae}^{- 4 t} + {Be}^{3 t} + C = - 3 {Ae}^{- 4 t} + 4 {Be}^{3 t} + C = \frac{- 3 {Ae}^{- 4 t} + 4 {Be}^{3 t} + C}{6} = - \frac{1}{2} {Ae}^{- 4 t} + \frac{2}{3} {Be}^{3 t} + \frac{C}{6} x (t) = - \frac{1}{2} {Ae}^{- 4 t} + \frac{2}{3} {Be}^{3 t} + \frac{C}{6} \dots (9)$

Now substitute the equation (8) and (9) in equation (4)

$(D - 1) [x] - 2 [y] - [z] = 0 z = (D - 1) [x] - 2 [y] z = (D - 1) [- \frac{1}{2} {Ae}^{- 4 t} + \frac{2}{3} {Be}^{3 t} + \frac{C}{6}] - 2 [{Ae}^{- 4 t} + {Be}^{3 t} + C] = \frac{4}{2} {Ae}^{- 4 t} + \frac{6}{3} {Be}^{3 t} + \frac{1}{2} {Ae}^{- 4 t} - \frac{2}{3} {Be}^{3 t} - \frac{C}{6} - 2 {Ae}^{- 4 t} - 2 {Be}^{3 t} - 2 C = 2 {Ae}^{- 4 t} + 2 {Be}^{3 t} + \frac{1}{2} {Ae}^{- 4 t} - \frac{2}{3} {Be}^{3 t} - \frac{C}{6} - 2 {Ae}^{- 4 t} - 2 {Be}^{3 t} - 2 C = \frac{1}{2} {Ae}^{- 4 t} - \frac{2}{3} {Be}^{3 t} - \frac{13}{6} C$

So, the solution is founded.

Most popular questions for Math Textbooks

Suppose the coupled mass-spring system of Problem (Figure 5.31) is hung vertically from support (with mass $m_{2}$ above $m_{1}$ ), as in Section 4.10, page 226.

(a) Argue that at equilibrium, the lower spring is stretched a distance $l_{1}$ from its natural length $l_{2}$ , given by $l_{1} = m_{1} g / k_{1}$ .

(b) Argue that at equilibrium, the upper spring is stretched a distance $l_{2} = (m_{1} + m_{2}) g / k_{2}$ .

(c) Show that if $x_{1}$ and $x_{2}$ are redefined to be displacements from the equilibrium positions of the masses $m_{1}$ and $m_{2}$ , then the equations of motion are identical with those derived in Problem 1.

In Problems 3 – 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.

$\frac{dx}{dt} = x - 4 y \frac{dy}{dt} = x + y$

In Problems 15–18, find all critical points for the given system. Then use a software package to sketch the direction field in the phase plane and from this description the stability of the critical points (i.e., compare with Figure 5.12).

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

In Problems 25 – 28, use the elimination method to find a general solution for the given system of three equations in the three unknown functions x(t), y(t), z(t).

$x' = 3 x + y - z, y' = x + 2 y - z, z' = 3 x + 3 y - z$

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].

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