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.
determine y (0
2) with the fourth-order Taylor series methodusing a single integration
step. Also compute the estimated error fromEq. (7.7) and compare it with the actual
error. The analytical solution of the differentialequationis
1
32
Solution The Taylor series up to and including the termwith h 4 is
31
32 e 4 x
1
4 x 2
1
8 x
=
+
+
y
1
2! y (0) h 2
1
3! y (0) h 3
1
4! y (4) (0) h 4
y (0) h
y ( h )
=
y (0)
+
+
+
+
(a)
Differentiation of the differentialequationyields
y =−
x 2
4 y
+
y =−
4 y +
4 x 2
2 x
=
16 y
+
2 x
y =
16 y
16 x 2
8 x
+
2
=−
64 y
+
8 x
+
2
y (4)
64 y +
64 x 2
=−
32 x
8
=
256 y
+
32 x
8
Thus
y (0)
=−
4(1)
=−
4
y (0)
=
16(1)
=
16
y (0)
=−
64(1)
+
2
=−
62
y (4) (0)
=
256(1)
8
=
248
With h
=
0
.
2Eq. (a) becomes
1
2! (16)(0
1
3! (
1
4! (248)(0
2) 2
2) 3
2) 4
y (0
.
2)
=
1
+
(
4)(0
.
2)
+
.
+
62)(0
.
+
.
=
.
0
4539
According to Eq. (7.7) the approximate truncation erroris
h 4
5! y (4) (0
y (4) (0)
E
=
.
2)
where
y (4) (0)
=
248
y (4) (0
2) 2
.
2)
=
256(0
.
4539)
64(0
.
+
32(0
.
2)
8
=
112
.
04
Therefore,
.
2) 4
5!
(0
E
=
(112
.
04
248)
=−
0
.
0018
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