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Table 5 Differential transforms G i (k) for g(t)=y(t)ln[y(t)] , k=0,1,2,3,4

From: A semi-analytic method with an effect of memory for solving fractional differential equations

k G i (k)
0 Y i (0)ln[ Y i (0)]
1 Y i (1)ln[ Y i (0)]+ Y i (1)
2 Y i (2)ln[ Y i (0)]+ Y i (2)+ ( Y i ( 1 ) ) 2 2 Y i ( 0 )
3 Y i (3)ln[ Y i (0)]+ Y i (3)+ Y i ( 1 ) Y i ( 2 ) Y i ( 0 ) ( Y i ( 1 ) ) 3 6 ( Y i ( 0 ) ) 2
4 Y i (4)ln[ Y i (0)]+ Y i (4)+ 2 Y i ( 1 ) Y i ( 3 ) + ( Y i ( 2 ) ) 2 2 Y i ( 0 ) ( Y i ( 0 ) ) 2 Y i ( 2 ) 2 ( Y i ( 0 ) ) 2 + ( Y i ( 1 ) ) 4 12 ( Y i ( 0 ) ) 3