Métodos numericos: ecuaciones diferenciales ordinarias

de donde el desarrollo de Taylor de orden dos es
G2(h) = G2(0) + G 0 2(0)h + 1
2 G00 2(0)h 2 + O(h 2 )
µ
∂
y por otra parte
que derivando dos veces nos da
G 0 3(h) =
∂
c3
G 00
3(h) = c 2 3
de donde
G 00
= f(ti−1, yi−1)+ c2
∂t f(ti−1, yi−1)+a21f(ti−1, yi−1) ∂
∂y f(ti−1, yi−1)
+ 1
µ
c
2
2 ∂
2
2
∂t2 f(ti−1, yi−1)+2c2a21f(ti−1, yi−1) ∂2
∂t∂y f(ti−1, yi−1)
+ a 2 2 ∂2
21f(ti−1, yi−1)
∂y2 f(ti−1,
yi−1) h 2 + O(h 3 )
G3(h) =f(ti−1 + c3h, yi−1 + a31g1 + a32hG2(h)),
Métodosdeunpaso
∂t f(ti−1 + c3h, yi−1 + a31g1 + a32hG2(h))
+ ∂
∂y f(ti−1 + c3h, yi−1 + a31g1 + a32hG2(h)) · (a31f(ti−1, yi−1)+a32(G2(h)+hG 0 2(h))) ,
∂2 ∂t2 f(ti−1 + c3h, yi−1 + a31g1 + a32hG2(h)) +
∂2 ∂t∂y f(ti−1 + c3h, yi−1 + a31g1 + a32hG2(h)) · (a31f(ti−1, yi−1)+a32(G2(h)+hG 0 2(h)))
+2c3
+ ∂2
∂y 2 f(ti−1 + c3h, yi−1 + a31g1 + a32hG2(h)) · (a31f(ti−1, yi−1)+a32(G2(h)+hG 0 2(h))) 2
+ ∂
∂y f(ti−1 + c3h, yi−1 + a31g1 + a32hG2(h)) (a32(2G 0 2(h)+hG 00
2(h))) ,
3(0) = c 2 3
G 0 ∂
3(0) = c3
∂t f(ti−1, yi−1)+(a31 + a32) ∂
∂2 ∂t2 f(ti−1, yi−1)+2c3 (a31 + a32) ∂2
+(a31 + a32) 2 ∂ 2
+2a32
∂
∂y f(ti−1, yi−1)
∂y 2 f(ti−1, yi−1)f(ti−1, yi−1) 2
por lo que el desarrollo de Taylor en cero es
G3(h) = G3(0) + G 0 3(0)h + 1
2 G00 3(0)h 2 + O(h 3 )
µ
∂
= f(ti−1, yi−1)+
µ
∂
∂y f(ti−1, yi−1)f(ti−1, yi−1),
∂t∂y f(ti−1, yi−1)f(ti−1, yi−1)
h
c2
∂t f(ti−1, yi−1)+a21f(ti−1, yi−1) ∂
∂y f(ti−1, yi−1)
c3
∂t f(ti−1, yi−1)+(a31 + a32) ∂
41
,
∂y f(ti−1,
yi−1)f(ti−1, yi−1) h