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Verify that the solution to the initial value problem

x'=5x-3y-2;    x0=2,y'=4x-3y-1;   y0=0

Satisfies |xt|+|yt|→+∞ast→+∞

Short Answer

Expert verified

The solutions for the given initial value problem are xt=-54e-t+94c2e3t+1

and yt=-52e-t+32e3t+1. Then, it satisfies the limt→∞xt+yt=∞also.

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, andL4 are polynomials in D=ddt:

  1. Make sure that the system is written in operator form.
  1. 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.
  1. (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.
  1. Eliminate x from the system and solve for y(t). [Solving for y(t) gives more constants----in fact, twice as many as needed.]
  1. 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,

x'=5x-3y-2      ......(1)y'=4x-3y-1       ......(2)

To verify:xt+yt→+∞ as t→+∞.

Let us rewrite this system of operators in operator form:

D-5x+3y=-2      ......(3)4x-D+3y=1         ......(4)

Multiply 4 on equation (3) and D-5 on equation (4). Then, add them together to get,

4D-5x+12y-4D-5x-D-5D+3y=-8-D-51D-5D+3y-12y=-8+5D2-2D-15+12y=-3D2-2D-3y=-3

D2-2D-3y=-3      ......(5)

Since the corresponding auxiliary equation is r2-2r-3=0. The roots are r=-1and r=3.

Then, the homogeneous solution isyht=c1e-t+c2e3t      ......(6)

Let us take the undetermined coefficients and assume thatypt=C      ......(7)

Now derivate the equation (7)

Dypt=0

Substitute the derivation in equation (5).

D2-2D-3C=-30-0-3C=-3C=1

So, ypt=1.

Then,yt=c1e-t+c2e3t+1      ......(8)

03

Substitution method

Substitute equation (8) in equation (4).

4x-D+3y=14x=1+D+3y4x=1+D+3c1e-t+c2e3t+1

4x=1-c1e-t+3c2e3t+3c1e-t+3c2e3t+3=2c1e-t+6c2e3t+4x=2c1e-t+6c2e3t+44=12c1e-t+32c2e3t+1

xt=12c1e-t+32c2e3t+1      ......(9)

04

Find the initial value problem

Given,    x0=2,   y0=0.

Now substitute the values in equations (8) and (9).

xt=12c1e-t+32c2e3t+1x0=12c1e0+32c2e30+11=12c1+32c2c1+3c2=2           ......(10)

yt=c1e-t+c2e3t+1y0=c1e-0+c2e30+1-1=c1+c2c1+c2=-1          ......(11)

First, solve the equations (10) and (11).

c1+3c2-c1-c2=2+12c2=3c2=32

Then,

3c1+3c2-c1-3c2=-3-22c1=-5c1=-52

Now substitute the values of c in equations (8) and (9).

xt=-54e-t+94c2e3t+1yt=-52e-t+32e3t+1

Now calculate the limits:

limt→∞xt+yt=limt→∞-54e-t+94c2e3t+1+-52e-t+32e3t+1=limt→∞-54e-t+94c2e3t+1+limt→∞-52e-t+32e3t+1=-0+∞+1+-0+∞+1limt→∞xt+yt=∞

So, the solution is founded.

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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).

Rigid Body Nutation. Euler’s equations describe the motion of the principal-axis components of the angular velocity of a freely rotating rigid body (such as a space station), as seen by an observer rotating with the body (the astronauts, for example). This motion is called nutation. If the angular velocity components are denoted by x, y, and z, then an example of Euler’s equations is the three-dimensional autonomous system

\(\begin{array}{l}\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ = yz}}\\\frac{{{\bf{dy}}}}{{{\bf{dt}}}}{\bf{ = - 2xz}}\\\frac{{{\bf{dz}}}}{{{\bf{dt}}}}{\bf{ = xy}}\end{array}\)

The trajectory of a solution x(t),y(t), z(t) to these equations is the curve generated by the points (x(t), y(t), z(t) ) in xyz-phase space as t varies over an interval I.

(a) Show that each trajectory of this system lies on the surface of a (possibly degenerate) sphere centered at the origin (0, 0, 0).[Hint: Compute\(\frac{{\bf{d}}}{{{\bf{dt}}}}{\bf{(}}{{\bf{x}}^{\bf{2}}}{\bf{ + }}{{\bf{y}}^{\bf{2}}}{\bf{ + }}{{\bf{z}}^{\bf{2}}}{\bf{)}}\)What does this say about the magnitude of the angular velocity vector?

(b) Find all the critical points of the system, i.e., all points\({\bf{(}}{{\bf{x}}_{\bf{o}}}{\bf{,}}{{\bf{y}}_{\bf{o}}}{\bf{,}}{{\bf{z}}_{\bf{o}}}{\bf{)}}\) such that \({\bf{x(t) = }}{{\bf{x}}_{\bf{o}}}{\bf{,y(t) = }}{{\bf{y}}_{\bf{o}}}{\bf{,z(t) = }}{{\bf{z}}_{\bf{o}}}\) is a solution. For such solutions, the angular velocity vector remains constant in the body system.

(c) Show that the trajectories of the system lie along the intersection of a sphere and an elliptic cylinder of the form\({{\bf{y}}^{\bf{2}}}{\bf{ + 2}}{{\bf{x}}^{\bf{2}}}{\bf{ = C}}\) for some constant C. [Hint: Consider the expression for dy/dx implied by Euler’s equations.]

(d) Using the results of parts (b) and (c), argue that the trajectories of this system are closed curves. What does this say about the corresponding solutions?

(e) Figure 5.19 displays some typical trajectories for this system. Discuss the stability of the three critical points indicated on the positive axes.

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.

dxdt+y=t2,-x+dydt=1

Show that the Poincare map for equation (1) is not chaoticby showing that if\({\bf{(}}{{\bf{x}}_{\bf{o}}}{\bf{,}}{{\bf{\nu }}_{\bf{o}}}{\bf{)}}\)and\({\bf{(}}{{\bf{x}}^{\bf{*}}}_{\bf{o}}{\bf{,}}{{\bf{\nu }}^{\bf{*}}}_{\bf{o}}{\bf{)}}\)are two initial values that define the Poincare maps\({\bf{(}}{{\bf{x}}_{\bf{n}}}{\bf{,}}{{\bf{\nu }}_{\bf{n}}}{\bf{)}}\) and\({\bf{(}}{{\bf{x}}^{\bf{*}}}_{\bf{n}}{\bf{,}}{{\bf{\nu }}^{\bf{*}}}_{\bf{n}}{\bf{)}}\), respectively, using the recursive formulas in (3), then one can make the distance between\({\bf{(}}{{\bf{x}}_{\bf{n}}}{\bf{,}}{{\bf{\nu }}_{\bf{n}}}{\bf{)}}\)and\({\bf{(}}{{\bf{x}}^{\bf{*}}}_{\bf{n}}{\bf{,}}{{\bf{\nu }}^{\bf{*}}}_{\bf{n}}{\bf{)}}\)small by making the distance between\({\bf{(}}{{\bf{x}}_{\bf{o}}}{\bf{,}}{{\bf{\nu }}_{\bf{o}}}{\bf{)}}\) and \({\bf{(}}{{\bf{x}}^{\bf{*}}}_{\bf{o}}{\bf{,}}{{\bf{\nu }}^{\bf{*}}}_{\bf{o}}{\bf{)}}\)small. (Hint: Let \({\bf{(A,}}\phi {\bf{)}}\)and \({\bf{(}}{{\bf{A}}^{\bf{*}}}{\bf{,}}{\phi ^ * }{\bf{)}}\) be the polar coordinates of two points in the plane. From the law of cosines, it follows that the distance d between them is given by\({{\bf{d}}^{\bf{2}}}{\bf{ = (A - }}{{\bf{A}}^{\bf{*}}}{{\bf{)}}^{\bf{2}}}{\bf{ + 2A}}{{\bf{A}}^{\bf{*}}}{\bf{(1 - cos(}}\phi {\bf{ - }}{\phi ^ * }{\bf{))}}\).)

Find all the critical points of the system

dxdt=x2-1dydt=xy

and the solution curves for the related phase plane differential equation. Thereby proving that two trajectories lie on semicircles. What are the endpoints of the semicircles?

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