Application of the Multistep Generalized Differential Transform Method to Solve a Time-Fractional Enzyme Kinetics

Table 1

Operations of differential transformation.

Original function	Transformed function
𝑓(𝑡)=𝑢(𝑡)±𝑣(𝑡)	𝐹(𝑘)=𝑈(𝑘)±𝑉(𝑘)
𝑓(𝑡)=𝛼𝑢(𝑡)	𝐹(𝑘)=𝛼𝑈(𝑘)
𝑓(𝑡)=𝑢(𝑡)𝑣(𝑡)	𝐹(𝑘)=∑^𝑘𝑙=0𝑈(𝑙)𝑉(𝑘−𝑙)
𝑓(𝑡)=𝑑𝑢(𝑡)𝑑𝑡	𝐹(𝑘)=(𝑘+1)𝑈(𝑘+1)
𝑓(𝑡)=𝑑^𝑚𝑢(𝑡)𝑑𝑡^𝑚	𝐹(𝑘)=(𝑘+1)(𝑘+2)⋯(𝑘+𝑚)𝑈(𝑘+𝑚)
𝑓(𝑡)=∫^𝑡𝑡0𝑢(𝑡)𝑑𝑡	𝐹(𝑘)=𝑈(𝑘−1)𝑘,𝑘≥1
𝑓(𝑡)=𝑡^𝑚	𝐹(𝑘)=𝛿(𝑘−𝑚)=1,𝑘=𝑚0,𝑘≠𝑚
𝑓(𝑡)=exp(𝜆𝑡)	𝐹(𝑘)=𝜆^𝑘𝑘!
𝑓(𝑡)=sin(𝜔𝑡+𝛼)	𝐹(𝑘)=𝜔^𝑘𝑘!sin𝜋𝑘2+𝛼
𝑓(𝑡)=cos(𝜔𝑡+𝛼)	𝐹(𝑘)=𝜔^𝑘𝑘!cos𝜋𝑘2+𝛼
ℎ(𝑡)=𝑓(𝑡)𝑔(𝑡)	𝐻(𝑘)=1𝐺(0)𝐹(𝑘)−∑^𝑘−1𝑚=0𝐻(𝑚)𝐺(𝑘−𝑚)
𝑓(𝑡)=𝑔(𝑡)^𝑏	𝐹(𝑘)=𝐺(0),𝑘=0𝑘𝑚∑=1(𝑏+1)𝑚−𝑘(0)𝐺(𝑚)𝐹()𝑘𝐺𝑘−𝑚,𝑘≥1