FOR DIFFERENTIAL EQUATIONS OF FRACTIONAL ORDER

Boundary Value Problems for Differential Equations of Fractional Order 59Since the solution (2.37) is an entire function of the parameter λ, we have∞∑u(x, λ) = S n (x)λ n . (2.38)Substituting (2.38) in (2.37) we have= x +∫ x0From (2.39) it followsn=0S 0 (x) + λS 1 (x) + λ 2 S 2 (x) + · · · =[(x − t) 1−α − λ(x − t) ][ S 0 (t) + · · · + λ n S n (t) + · · · ] dt. (2.39)S 0 (x) = x +S 1 (x) = −S 2 (x) =∫ x0∫ x0∫ x0(x − t) 1−α u(t) dt, (2.40)(x − t)S 0 (t) dt +∫ x(x − t) 1−α S 2 (x) dt +0∫ x(x − t) 1−α S 1 (x) dt, (2.41)0(x − t) 1−α S 1 (x) dt, (2.42). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Solving the equation (2.40), we obtainS 0 (x) = xE ρ (x 1/ρ ; 2); S 0 (1) = 1.For solution of the equation (2.41) we will calculateThen∫ x0(x − t)S 0 (t) dt =a 0 (x)=x 3 E ρ (x 1/ρ ; 4)+∫ x0∫ x0tE ρ (t 1/ρ ; 2)(x − t) dt = x 3 E ρ (x 1/ρ ; 4).((x − t) 1/ρ−1 E ρ (x−t) 1/ρ ; 1 )t 3 E ρ (t 1/ρ ; 4) dt =ρ=−c 1(x 3 E ρ (x 1/ρ ; 4)+c 0 ρx 3+1/ρ[ E ρ(x 1/ρ ; 3+ 1 ρIt is likewise possible to show that(− 3+ 1 ( )E ρ x 1/ρ ; 4+ 1 ρρ)] ) .a 2 (x) = −cx 5 E ρ (x 1/ρ ; 6) + cρx 5+1/ρ E ρ(x 1/ρ ; 5 + 1 ρ)−(− cρ 5 + 1 ()x 5+1/ρ E ρ x 1/ρ ; 6 + 1 )−ρρ)−