1 SISTEMAS DE ECUACIONES DIFERENCIALES ORDINARIAS

ITERACIÓN 1
k 1 = f (t 0 , x 0 , y 0 ) = f (0, -1, 6) = 2(-1) + 4(6) = 22
m 1 = g (t 0 , x 0 , y 0 ) = g (0; -1; 6) = -(-1) + 6(6) = 37
h h
k 2
= f (t 0
+ , x 0 + k 1 , y 0 + hm 1 ) = f(0.1;1.2;9.7)= 41.2
2 2
h h
2
m 2
= g (t 0
+ , x 0 + k 1 , y 0 + h m 1 ) = g(0.1;1.2;9.7)= 57
2 2
2
k 3
= f (t 0
+
h
, x 0 +
h
k 1 , y 0 + hm 1 ) = f(0.1;3.12; 11.7 )= 53.04
2 2
2
m 3
= g (t 0
+
h
, x 0 +
h
k 1 , y 0 + h m 1 ) = g(0.1;3.12; 11.7 )= 67.08
2 2
2
k 4
= f (t 0
+ h , x 0 + h k 1 , y 0 + h m 1 ) = f(0.2; 9.608; 19.416 )= 96.88
m 4
= g (t 0
+ h , x 0 + h k 1 , y 0 + h m 1 ) = g(0.2; 9.608; 19.416 )= 104.08
h
x 1
= x 0
+ (k 1 + 2k 2 +2k 3 + k 4 )= 9.2453
6
h
y 1
= y 0
+ (m 1 + 2m 2 +2m 3 + m 4 )= 18.9747
6
t 1 = t 0 + h = 0+0.2=0.2
9
Así sucesivamente, obtenemos la tabla siguiente:
t
x n
y n
0
-1.0
6.0
0.2
k 1
=22
m 1
=37
k 2
=41.2
m 2
=57
k 3
=10.6080
m 3
=13.4160
k 4
=19.3760
m 4
=21.3776
9.2453
18.9747
0.4
k 1
=94.3894
m 1
=104.6029
k 2
=155.1084
m 2
=157.9257
k 3
=188.5814
m 3
=183.8475
k 4
=316.8999
m 4
=288.4504
45.8676
54.8614
0.6
k 1
=311.1808
m 1
=283.3008
k 2
=486.7373
m 2
=422.1632
k 3
=577.3935
m 3
=487.9250
k 4
=932.4782
m 4
=771.2258
158.2650
150.6848
10
5