Capitolul 1 Ecuatii diferentiale de ordinul ˆıntâi rezolvabile prin ...

138 CAPITOLUL 4
> h:=0.1: n:=10:
> f1:=(t,x1,x2)->-x1(t)+8*x2(t): f2:=(t,x1,x2)->x1(t)+x2(t):
> t:=(n,h)->n*h:
> x1:=proc(n,h);
> if n=0 then x1(0) else
x1(n-1,h)+h*f1(t(n-1,h),x1(n-1,h),x2(n-1,h))end if;
> end proc:
> x2:=proc(n,h)
> if n=0 then x2(0)else
x2(n-1,h)+h*f2(t(n-1,h),x1(n-1,h),x2(n-1,h))end if;
> end proc:
> x1(0):=1: x2(0):=1:
> x1(t):=[seq(x1(i,h),i=0..n)];
x1 (t) := [1, 1.7, 2.49, 3.433, 4.6001, 6.07617, 7.966249,
10.4031833, 13.55708001, 17.64726322, 22.95758363]
> x2(t):=[seq(x2(i,h),i=0..n)];
x2 (t) := [1, 1.2, 1.49, 1.888, 2.4201, 3.12212, 4.041949,
> t:=[seq(t(i,h),i=0..n)];
5.2427688, 6.80736401, 8.843808412,11.49291558]
t := [0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0]
Pentru a compara valorile numerice ale solut¸iei cu cele obt¸inute
prin calcul simbolic, vom calcula valorile solut¸iei obt¸inute în Capitolul 4
în câteva puncte:
> sol_x1:=5/3*exp(3*t)-2/3*exp(-3*t):
> sol_x2:=5/6*exp(3*t)+1/6*exp(-3*t):
> eval(sol_x1(t),t=0);
eval(sol_x1(t),t=0.1);eval(sol_x1(t),t=0.2);