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2
((
((
dy t n)) dy t n)) (∆ t) 2
((
((
ytn + 1 )) = ytn)) + (∆ t) + (4.45)
dt dt 2 2
Recalling Eq. (4.41) and the total derivative expression of a function in two
variables as function of the partial derivatives, we have:
dy t n(( ))
= ft n y t n( ( ), ( ( ))) (4.46)
dt

2
dy t n(( )) = d   dy t n(( )) 
dt 2 dt  dt 
(4.47)
∂ ft n y t n ( ))) ∂ ft n y t n ( )))
( ( ), (
( ( ), (
= + ft n y t n ( )))
( ( ), (
t ∂ y ∂
Combining Eqs. (4.45) to (4.47), it follows that to second order in (∆t):

ytn( ( + 1 )) = ytn( ( )) + f tn ytn( ( ), ( ( )))(∆ t) +
∂ ft n y t n( ( ), ( ( ))) ∂ ft n y t n( ( ), ( ( ))) (  ∆ t) 2 (4.48)
+  + ft n y t n( ( ), ( ( ))) 
 t ∂ y ∂  2

Next, let us Taylor expand k to second order in (∆t). This results in:
2
t =
f tn +
∆
k = ( ( ) α∆ t y tn + β k )( )
, ( ( ))
2 1
 ∂ ft n y t n ( ))) ∂ ft n y t n ( )))  (4.49)
( ( ), (
( ( ), (
 ft n y t n ( ))) + (α ∆ t) + (β k )  (∆ t)
( ( ), (
1
 t ∂ y ∂ 
Combining Eqs. (4.42), (4.44), and (4.49), we get the other expression for
y(t(n + 1)), correct to second order in (∆t):
ytn( ( + 1 )) = ytn( ( )) (+ a b f tn ytn) ( ( ), ( ( )))(+ ∆ t) +
∂ ft n y t n( ( ), ( ( ))) ∂ ft n y t n( ( ), ( ( ))) (4.50)
+ α b (∆ t) + bβ ft n y t n( ( ), ( ( )))(∆ t) 2
2
t ∂ y ∂
Now, comparing Eqs. (4.48) and (4.50), we obtain the following equalities:

ab+= 1; α b = 1 2/ ; b = 1β (4.51)

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