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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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