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required for the calculation of
k
4
. Thus,
h
∂
2
h
∂
h
∂
f
1
2!
k
3
∂
∂y
1
2
k
3
∂
∂y
k
3
∂
f
∂y
∂
r
+
f
=
∂
r
+
∂
r
+
h
2
∂
2
f
∂
r
+
2
f
∂
r
∂y
+
2
f
∂y
1
2
2
hk
3
∂
k
3
∂
=
2
1
2
h
2
∂
2
h
hf
2
Df
∂
2
f
∂
r
2
+
h
2
2
f
∂
r
∂y
=
+
r
=
r
i
+
h
2
f
2
h
3
fDf
∂
2
f
∂y
O
h
4
+
+
2
r
=
r
i
+
h
2
2
Df
∂
h
3
2
f
∂
r
∂y
+
2
f
∂y
O
h
4
f
∂
2
D
2
f
=
+
2
r
=
r
i
+
h
2
h
3
2
Df Df
y
O
h
4
,
2
D
2
f
=
+
(3.291)
again using (3.268) and the identity
2
f
∂
r
∂y
+
2
f
∂y
∂
f
i
∂
2
=
Df
y
.
(3.292)
Again,
h
∂
3
h
∂
h
2
∂
2
f
∂
r
2
+
2
f
∂
r
∂y
+
2
f
∂y
1
3!
k
3
∂
∂y
1
6
k
3
∂
∂y
2
hk
3
∂
k
3
∂
∂
r
+
f
=
∂
r
+
2
r
=
r
i
+
h
3
6
∂
∂
r
+
∂
2
f
∂
r
2
+
2
f
∂
r
∂y
+
2
f
∂y
f
∂
∂y
2
f
∂
f
2
∂
O
h
4
=
2
6
D
3
f
r
=
r
i
+
h
3
O
h
4
,
=
(3.293)
using (3.270). From (3.290) with substitution from (3.289), (3.291) and (3.293),
we find that
k
4
=
hf
(
r
i
+
h
,y
i
+
k
3
)
hf
2
D
2
f
f
y
Df
h
3
h
2
Df
=
+
+
+
r
=
r
i
+
D
3
f
h
4
6
3
3
O
h
5
.
4
f
y
D
2
f
2
f
2
+
+
+
y
Df
+
3
Df Df
y
(3.294)
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