Page 16-65
Press
@SOLVE
(wait)
@EDIT
to solve for w(t=2). The solution reads [.16716… -
.6271…], i.e., x(2) = 0.16716, and x'(2) = v(2) = -0.6271. Press
@CANCL
to
return to SOLVE environment.
Solution presented as a table of values
In the previous example we were interested only in finding the values of the
position and velocity at a given time t. If we wanted to produce a table of
values of x and x', for t = 0.00, 0.25, …, 2.00, we will proceed as follows:
First, prepare a table to write down your results:
Next, within the SOLVE environment, change the final value of the
independent variable to 0.25, use:
—
.25
@@OK@@
™™
@SOLVE
(wait)
@EDIT
(Solves for w at t = 0.25, w = [0.968 1.368]. )
@@OK@@
INIT+
—
. 5
@@OK@@
™™
@SOLVE
(wait)
@EDIT
(Changes initial value of t to 0.25, and final value of t to 0.5, solve again for
w(0.5) = [0.748 -2.616])
@@OK@@ @INIT+
—
.75
@@OK@@
™™
@SOLVE
(wait)
@EDIT
(Changes initial value of t to 0.5, and final value of t to 0.75, solve again for
w(0.75) = [0.0147 -2.859])
@@OK@@ @INIT+
—
1
@@OK@@
™ ™
@SOLVE
(wait)
@EDIT
(Changes initial value of t to 0.75, and final value of t to 1, solve again for
w(1) = [-0.469 -0.607])
t x x'
0.00 0.00 6.00
0.25
… … …
2.00
