HP 50G, User Manual

Page: 539 / 887

Page 16-62
time t = 2, the input form for the differential equation solver should look as
follows (notice that the Init: value for the Soln: is a vector [0, 6]):
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
t x x'
0.00 0.00 6.00
0.25
… … …
2.00