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On the solution of some fractional differential equations

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Fig. 3 (A)                                                     Fig. (B)  ( ) 4 3/ 2 6 3 e erf ( ) 3tt i t i ty tπ π−−−=               ( ) 0t e erfu ( ) sin ( )y t= −i∫−i ut− u du Case 3: If  3α= 2 , 1β=2 c 0 = 1, c 1 = 2 and 0f t( )=  then  y t ( ) = − 4 4

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