1550078481-Ordinary_Differential_Equations__Roberts_

PORTRAIT User's Guide 533

Substituting this expression for y into the second equatio n of (5), we see x

must satisfy

2 2 1 2

8x-(

3

x +

3

) -7=0.

Squaring, multiplying the resulting equation by 9, and rearranging, we see x

must satisfy the quartic equation

4x^4 + 4x^2 - 72x + 64 = 0.

Using POLYRTS to solve this equation, we find x = 1, x = 2, and x = -1.5±

2.397916i. Substituting x = 1 into equation (6), yields y = 1. Hence, one real
critical point of system (4) is (1, 1). Substituting x = 2 into equation (6), we
find y = 3. Therefore, the second real crit ical point of system (4) is (2, 3).
Based on these two real critical points, we decided to produce the phase-plane
graph on the rectangle R = {(x, y) I 0 :::; x:::; 5 and 0 :::; y:::; 5}.
First, we chose the point (0, 2.5) on the left edge of the rectangle R and
generated a numerical solution to the IVP consisting to the system ( 4) and
the initial condition (i) x(O) = 0, y(O) = 2.5. At first, we selected the interval
of integration to be t he interval [O, 2 .5]. We displayed the phase-plane graph
(trajectory) of this solut ion and saw that it left the rectangle near (0.5, 0). By
displaying the solution values, we saw that y became negative at time 1.135.
Hence, we reran this IVP and changed the interval of integration to [ 0, 1.1].
The resulting phase-plane graph is shown in Figure B. 7. As time increases,
this trajectory, which starts at the point (0, 2.5) at time t = 0, moves in a
general direction toward the critical point at (1, 1) and then exits the rectangle
R near (0.5, 0).

• D x
  Enter the number of initinl v alue problems to •olve

and Fo1 each tho initinitiaial l conditions value probld1 e m i and i. enter d2i. tho inte rval of integration (ai. bi]. the initia l value ci.

1.1

d2i -

2.5

Horizontal R a nge

HmHmax in-_,__~--r

Vmin Hmin -,...,. __ .... ;_~-'---..._..._...;.~..,;,.,-~n--.-..,.,.... Hmax

F igure B.7 Phase-Plane Graph of System (4) Using Initial Conditions (i)