Table 2:
Comparison of the numerical solution and error obtained by present method with series solution [
25
].
𝑥
Our 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
.
0
5
𝐸
−
0
5
2
.
5
3
𝐸
−
0
6
6
.
3
2
𝐸
−
0
7
5
.
8
5
𝐸
−
0
7
0.2
4
.
0
0
𝐸
−
0
5
2
.
5
0
𝐸
−
0
6
6
.
2
6
𝐸
−
0
7
6
.
0
4
𝐸
−
0
7
0.5
3
.
6
5
𝐸
−
0
5
2
.
2
8
𝐸
−
0
6
5
.
7
2
𝐸
−
0
7
5
.
5
8
𝐸
−
0
7
1.0
2
.
5
1
𝐸
−
0
5
7
.
8
8
𝐸
−
0
7
5
.
0
7
𝐸
−
0
7
8
.
2
0
𝐸
−
0
7