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Chapter 5: Q15E (page 271)

In Problems 15鈥�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).

Short Answer

Expert verified

This is unstable saddle point is (-2,1).

Step by step solution

01

Find critical points

Here the system is;

dxdt=2x+y+3dydt=-3x-2y-4

For critical points equate the system equal to zero.

2x+y+3=0-3x-2y-4=0

Solve for x and y by eliminating the method.

The values of x=-2 and y=1.

So, this is the unstable saddle point is (-2,1).

02

Sketch

This is the required result.

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Most popular questions from this chapter

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

D-3x+D-1y=t,D+1x+D+4y=1

In Problems 7鈥�9, solve the related phase plane differential equation (2), page 263, for the given system.

dxdt=x2-2y-3,dydt=3x2-2xy

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

dxdt=-8y,dydt=18x

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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.
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