Computer Algebra Recipes

3.2. SECOND-ORDER MODELS 129
Using the sequence command, we graph the numerical solution from x =0 to
X in steps of size 1=d. Each numerical point is plotted as a size-16 blue circle.
> gr1:=plot([seq([i/d,Ynum(i/d)],i=0..d*X)],style=point,
symbol=circle,symbolsize=16,color=blue):
The series solution is plotted as a solid thick curve over the same range of x,
> gr2:=plot(Y[N],x=0..X,thickness=2):
and the two graphs superimposed to produce Figure 3.5.
> display(fgr1,gr2g,labels=["x","y"]);
40
30
y
20
10
0 1 2 x 3 4
Figure 3.5: Comparison of numerical (circles) and series (curve) solutions.
Of course, because of the imposition of the conditional statement, the ¯t of the
series solution to the numerical points is very good.
Vectoria hopes that Jennifer will like her solution to this problem. By
modifying Vectoria's recipe, you should be able to solve the following problems,
which also appeared on Jennifer's assignment.
PROBLEMS:
Problem 3-12: What's special about this ODE?
Consider the following linear second-order ODE:
xy 00 (x)+y 0 (x)+xy(x) =0; y(0) = 1;y 0 (0) = 0:
(a) Show that Maple is able to produce a closed-form analytic solution. Identify
the ODE and the \special" function that appears in the solution.
(b) Derive a series solution for y(x), keeping su±cient terms to achieve 1%
agreement with the analytic answer at x = X =10.
(c) Plot the series derived in part (b) and the analytic solution together in the
same ¯gure over the range x =0toX, representing the analytic answer
by appropriately sized and spaced circles.