Table 3:
Comparison of the numerical solution and error obtained by present method with series solution [
25
].
𝑥
Present method
Series method [
25
]
𝑚
=
3
2
𝑚
=
1
2
8
𝑚
=
2
5
6
0.0
0
.
0
0
𝐸
+
0
0
0
.
0
0
𝐸
+
0
0
0
.
0
0
𝐸
+
0
0
0
.
0
0
𝐸
+
0
0
0.1
4
.
6
6
𝐸
−
0
4
2
.
9
1
𝐸
−
0
5
7
.
2
8
𝐸
−
0
6
3
.
9
4
𝐸
−
0
6
0.5
1
.
4
6
𝐸
−
0
4
9
.
1
6
𝐸
−
0
6
2
.
2
9
𝐸
−
0
7
3
.
0
2
𝐸
−
0
6
1.0
8
.
6
0
𝐸
−
0
5
3
.
6
8
𝐸
−
0
6
1
.
2
8
𝐸
−
0
8
9
.
3
1
𝐸
−
0
7