Computer Algebra Recipes

112 CHAPTER 3. LINEAR ODE MODELS
case, the assumptions are determined by trial and error, i.e., looking at the
form of the answer, which is now given.
Methods for ¯rst order ODEs:
| Trying classi¯cation methods |
trying a quadrature
trying 1st order linear
< ¡t
e
( RC
>:
)μ
P + AC¼¿R2
¿ 2 + ¼ 2 R 2 C 2
AC¼¿R
¡
2 μ
¼t
cos
¿
¿ 2 + ¼2R2C 2 +
AR¿ 2 μ
¼t
sin
¿
¿ 2 + ¼2R2 ; t · ¿
C2 8
<
AC¼¿R
:
2 ¿¡t
e
( RC )
¿ 2 + ¼2R2 ¡t
+ e( RC
C2 )μ
P + AC¼¿R2
¿ 2 + ¼2R2C 2
; t · T
¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢
Intheabovesolution,wehaveshownonlytheanalyticform(thusthedots)
for the ¯rst cycle to save on text space. You can observe the complete solution
by executing the recipe on your computer. Because of the inclusion of the
infolevel[dsolve] command, we are told in the output that Maple recognizes
ode3 as a ¯rst-order linear ODE and presumably uses a standard method for
solving such an ODE. You may object to this \black box" nature of the dsolve
command, but obtaining the answer easily in our opinion is more important
than the mechanical details leading up to it. However, if the methodology is
important to you, we shall mimic a hand calculation in the next recipe.
To make a plot, we shall take the parameter values1 to be T = 1 second (60
heartbeats per minute), ¿ =0:15 s, C =0:002 liters/mm Hg, P =80mmHg,
and R = 1056 mm Hg/liter/second.
> tau:=0.15: T:=1: C:=0.002: P:=80: R:=1056:
The amplitude parameter A must be such that the right-hand side (rhs) of the
solution must be the same at t = T as at t = 0, i.e., the heartbeat is periodic.
> eq:=eval(rhs(sol),t=0)=eval(rhs(sol),t=T);
432:0593387 A¼
eq := 80: =
+49:82624166
0:0225 + 4:46054400 ¼2 The above equation is numerically solved for A.
> A:=fsolve(eq,A);
A := 0:9791418590
The pressure is then given by the rhs of sol, the °oating-point evaluation commandbeingusedtoexpresstheanswertofourdigits.
> Pressure:=evalf(rhs(sol),4);
1These approximate values are taken from the Internet.