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Chapter 5: Q12E (page 249)

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.

D2u+Dv=2,4u+Dv=6

Short Answer

Expert verified

The solutions for the given linear system areut=c1e-2t+c2e2t+1 and vt=2c1e-2t-2c2e2t+2t+c3.

Step by step solution

01

General form

Elimination Procedure for 2 × 2 Systems

To find a general solution for the system

L1x+L2y=f1,L3x+L4y=f2,

WhereL1,L2,L3 and L4are polynomials inD=ddt

a. Make sure that the system is written in operator form.

b. 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.

c. (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.

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

e. 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.

02

Evaluate the given equation

Given that,

D2u+Dv=2                  …1

4u+Dv=6                 …2

Then subtract equation (1) and (2) together one gets,

D2u+Dv-4u+Dv=2-6D2-4u=-4D2-4u=-4                      …3

Since the auxiliary equation to the corresponding homogeneous equation is r2-4=0. The roots are r=-2 and r=2.

Then, the homogeneous solution of u is;

uht=c1e-2t+c2e2t                    …4

Let us take the undetermined coefficients and assume that,

upt=1               …5

03

Substitution method

Use equations (4) and (5) to get,

ut=uht+uptut=c1e-2t+c2e2t+1                 …6

Now, take equation (2).

4u+Dv=6Dv=6-4c1e-2t+c2e2t+1Dv=-4c1e-2t-4c2e2t+2vt=2c1e-2t-2c2e2t+2t+c3

Thus, the solutions for the given linear system areut=c1e-2t+c2e2t+1 and vt=2c1e-2t-2c2e2t+2t+c3.

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Most popular questions from this chapter

The doubling modulo \({\bf{1}}\) map defined by the equation \(\left( {\bf{9}} \right)\)exhibits some fascinating behavior. Compute the sequence obtained when

  1. \({{\bf{x}}_0}{\bf{ = k / 7}}\)for\({\bf{k = 1,2, \ldots ,6}}\).
  2. \({{\bf{x}}_0}{\bf{ = k / 15}}\)for\({\bf{k = 1,2, \ldots ,14}}\).
  3. \({{\bf{x}}_{\bf{0}}}{\bf{ = k/}}{{\bf{2}}^{\bf{j}}}\), where \({\bf{j}}\)is a positive integer and \({\bf{k = 1,2, \ldots ,}}{{\bf{2}}^{\bf{j}}}{\bf{ - 1}}{\bf{.}}\)

Numbers of the form \({\bf{k/}}{{\bf{2}}^{\bf{j}}}\) are called dyadic numbers and are dense in \(\left( {{\bf{0,1}}} \right){\bf{.}}\)That is, there is a dyadic number arbitrarily close to any real number (rational or irrational).

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.

D+1u-D+1v=et,D-1u+2D+1v=5

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.

  1. Setting\({\bf{v = }}\frac{{{\bf{dx}}}}{{{\bf{dt}}}}\), convert the second-order equation to an equivalent first-order system.
  2. 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.
  3. 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.
  4. Physically, the case \({\bf{\lambda < 2}}\)corresponds to a current so high that the magnetic attraction completely overpowers the spring. To gain insight into this, use software to plot the phase plane diagrams for the system when \({\bf{\lambda = 1}}\) and when\({\bf{\lambda = 3}}\).
  5. 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.

Let A=D-1,B=D+2,C=D2+D-2, where D=ddt. For y=t3-8, compute

(a)Ay

(b)BAy

(c)By

(d)ABy

(e)Cy

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).

dxdt=-5x+2y,dydt=x-4y
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