Table 3: Comparison of the numerical solution and error obtained by present method with series solution [25].

π‘₯ Present method Series method [25]
π‘š = 3 2 π‘š = 1 2 8 π‘š = 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 βˆ’ 0 4 2 . 9 1 𝐸 βˆ’ 0 5 7 . 2 8 𝐸 βˆ’ 0 6 3 . 9 4 𝐸 βˆ’ 0 6
0.5 1 . 4 6 𝐸 βˆ’ 0 4 9 . 1 6 𝐸 βˆ’ 0 6 2 . 2 9 𝐸 βˆ’ 0 7 3 . 0 2 𝐸 βˆ’ 0 6
1.0 8 . 6 0 𝐸 βˆ’ 0 5 3 . 6 8 𝐸 βˆ’ 0 6 1 . 2 8 𝐸 βˆ’ 0 8 9 . 3 1 𝐸 βˆ’ 0 7