Fractional and operational calculus with generalized fractional ...

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$Numerical Methods Course Notes Version 0.1 (UCSD Math 174, Fall ...$

806 Ž. Tomovski et al. Downloaded By: [Srivastava, Hari M.] At: 18:19 27 October 2010 Theorem 6 The fractional differential equation (3.13) with the initial conditions (3.14) has its solution in the space L (0, ∞) given by ∞∑ (−1) m ∑ m ( ) m y (x) = a k b m−k x (α 3−α 2 )m+(α 2 −α 1 )k+α 3 −1 k m=0 + · c m+1 k=0 [ ac 1 x β 1(1−α 1 ) E α3 ,(α 3 −α 2 )m+(α 2 −α 1 )k+α 3 +β 1 (1−α 1 ) ∞∑ m=0 + bc 2 x β 2(1−α 2 ) E α3 ,(α 3 −α 2 )m+(α 2 −α 1 )k+α 3 +β 2 (1−α 2 ) + cc 3 x β 3(1−α 3 ) E α3 ,(α 3 −α 2 )m+(α 2 −α 1 )k+α 3 +β 3 (1−α 3 ) (−1) m c m+1 m ∑ k=0 ( − e ) c xα 3 ( − e ) c xα 3 ( − e ) ] c xα 3 ( ) m ( )( a k b m−k E m+1 α k 3 ,(α 3 −α 2 )m+(α 2 −α 1 )k+α 3 f − e ) c xα 3 . (3.15) Proof Making use of the above-demonstrated technique based upon the Laplace and the inverse Laplace transformations once again, it is not difficult to deduce the solution (3.15) just as we did in our proof of Theorem 5. 3.4. An Interesting Consequence of Theorem 6 Let 0
Integral Transforms and Special Functions 807 Remark 1 Podlubny [17] used the Laplace transform method in order to give an explicit solution for an arbitrary fractional linear ordinary differential equation with constant coefficients involving Riemann–Liouville fractional derivatives in series of multinomial Mittag–Leffler functions. 3.5. A Fractional Differential Equation with Variable Coefficient Kilbas et al. [14] used the Laplace transform method to derive an explicit solution for the following fractional differential equation with variable coefficients: x ( D α 0+ y) (x) = λy (x) ( x>0,λ∈ R; α>0; l − 1
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