Computer Algebra Recipes

3.2. SECOND-ORDER MODELS 127
On entering the given di®erential equation, Vectoria observes that the output
in de is expressed in terms of the prime notation.
> de:=2*x*diff(y(x),x,x)+(1-2*x)*diff(y(x),x)-x^2*y(x)=0;
de := 2 x y 00 +(1¡ 2 x) y 0 ¡ x2 y =0
The initial condition is speci¯ed, as well as the value of X and a parameter d
that will control the spacing of the numerical points in the graph.
> ic:=y(0)=1,D(y)(0)=0: X:=4: d:=8:
Using dsolve, an attempt is made to analytically solve de for y(x).
> dsolve(fde,icg,y(x)); #no analytic solution
No output is generated, Maple being unable to provide an analytic answer. Although
she has now omitted it, Vectoria had included infolevel[dsolve]:=5;
prior to the dsolve command. On doing so she was able to view a very lengthy
list of unsuccessful ODE solving methods tried by Maple. If you wish to see
these attempts, include the above infolevel command.
To generate a procedure for numerically evaluating y and y 0 at arbitrary x,
Vectoria includes the numeric option in dsolve and requests that the output
be given as a \list procedure."
> numsol:=dsolve(fde,icg,y(x),numeric,output=listprocedure);
numsol := [x =(proc(x) ::: end proc); y=(proc(x) ::: end proc);
y 0 =(proc(x) ::: end proc)]
Evaluating y(x) withthenumsol procedure will generate a numerical answer
for y at x when the value of x is supplied as the argument of Ynum de¯ned
below. Then, entering Ynum(X) gives the numerical value of y at X (4, here),
the answer being given to 18 digits, more than the \normal" 10-digit accuracy.
> Ynum:=eval(y(x),numsol): Ynum(X);
41:1065491999371986
Thus, y ¼ 41 at x = X = 4. A similar command structure is used to numerically
evaluate the derivative of y, yielding y 0 ¼ 76 at X.
> Ynumder:=eval(diff(y(x),x),numsol): Ynumder(X);
75:6306643796735756
Vectoria will use a do loop to systematically calculate series solutions of de as
a function of the order n, terms of order xn and larger being neglected. The
do loop will include a conditional statement that will terminate the loop when
the absolute percentage di®erence j100 (ynum(X)¡yseries(X))=ynum(X)j drops
below one percent. As a \seed number" to implement the conditional statement,
she calculates the absolute percentage di®erence between the starting value of
y(0) = 1 and the numerical value of y at x = X =4.
> percent[0]:=abs(evalf(100*(Ynum(X)-1)/Ynum(X))); #seed
percent 0 := 97:56729762
The do loop runs from 1 to 50, the number 50 being chosen to be large enough
to achieve a percentage error below one percent. The loop will calculate the