Culegere de probleme de Analiz˘a numeric˘a

166 Rezolvarea numerică ecuat¸iilor diferent¸iale
Solut¸ie. Predictorul cu m pas¸i se generează astfel:
xi+1
xi
(−1) k
y(xi+1) = y(xi)+
f(x,y(x))dx =
+ hm+1
m!
1
0
m−1
k=0
xi+1
xi
f(x,y(x))dx
∇ k f(xi,y(xi))h(−1) k
1
0
−s
s(s+1)...(s+m−1)f (m) (ξi,y(ξi))ds
k 0
s
ds 1
0 k
1
1
2
2
5
12
3
3
8
4
251
720
5
95
288
y(xi+1) = y(xi)+h f(xi,y(xi))+ 1
2 ∇f(xi,y(xi)+
1
Pentru m = 2 obt¸inem
+ 5
12 ∇2 f(xi,y(xi))+ 3
8 ∇3 f(xi,y(xi))+...
+h m+1 f (m) (µi,y(µi))(−1) m
1
y(xi+1) ≈ y(xi)+h
0
−s
ds
m
+
f(xi,y(xi))+ 1
2 ∇f(xi,y(xi))
=
k
ds+
= y(xi)+h f(xi,y(xi))+ 1
2 (f(xi,y(xi))−f(xi−1,y(xi−1)))
=
= y(xi)+ h
2 [3f(xi,y(xi))−f(xi−1,y(xi−1))]
h 3 f ′′ (µi,y(µi))(−1) 2
1
y0 = α, y1 = α1
yi+1 = yi + h
2 [3f(xi,yi)−f(xi−1,yi−1)]
−s
2
τi+1 = y(xi+1)−y(xi)
h
0
f ′′ (µi,y(µi)) = y (3) (µi)
ds = 5
12 h3 f ′′ (µi,y(µi))
− 1
2 [3f(xi,yi)−f(xi−1,yi−1)] =