# Solving a linear fourth-order fractional integro-differential equation

Definition: A real function $f(x)$, $x>0$ is said to be in space $C_\mu$, $\mu\in \Bbb R$ if there exists areal number $p$ ($>\mu$), such that $f(x)=x^pf_1(x)$, where $f_1(x)\in C[0,\infty]$, and it is said to be in the space $C^m_\mu$ iff $f^{(m)}\in C_\mu$, $m\in\Bbb N$. The Riemann-Liouville integral fractional operator of order $\alpha\geq 0$, of function $f\in C_\mu$, $\mu\geq -1$, is defined as $$J^\alpha f(x)=\frac{1}{\Gamma(\alpha)}\int_o^x(x-t)^{\alpha-1}f(t)dt,\quad \alpha>0,x>0,$$ $$J^0f(x)=f(x).$$ The fractional derivative of $f(x)$ in the Caputo sence is defined as $$D^\alpha_{*x}f(x)=\frac{1}{\Gamma(m-\alpha)}\int_0^x (x-t)^{m-\alpha-1}f^{(m)}(t) dt$$ for $m-1<\alpha\leq m$. $m\in\Bbb N$ and $f\in C^m_{-1}$.

I wanna solve the following linear forth-order fractional integro-differential equation $$D^\alpha_{*x}y(x)=x(1+e^x)+3e^x+y(x)-\int_0^x y(t)dt,\quad 0<x<1, 3<\alpha\leq 4,$$ subject to the boundary conditions $$y(0)=1,\quad y''(0)=2.$$ $$y(1)=1+e,\quad y''(1)=3e.$$ For instance, the exact solution, when $\alpha=4$ is $y(x)=1+xe^x$. The recursive Adomian decomposition algorithm is $$y_0=1+Ax+x^2+\frac{1}{3!}Bx^3+J^\alpha(x+xe^x+3e^x),$$ $$y_{n+1}=J^\alpha y_n(x)-J^\alpha\left(\int_0^x y_n(t)dt\right).\quad n\geq 0.$$ where the constants $A=y'(0)$ and $B=y'''(0)$ are to be determined. I don't know how to write the code.

a=4
DSolve[{1/Gamma[m - a]*
Integrate[(x - t)^(m - a - 1)*D[y[t], {t, m}], {t, 0, x}] ==
x (1 + Exp[x]) + 3 Exp[x] + y[x] - Integrate[y[t], {t, 0, x}],
y[0] == 1, y''[0] == 2, y[1] == 1 + E, y''[1] == 3 E}, y[x], {x,0,1}]

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Have you looked at any of the fractional calculus user code in the Wolfram library? –  Mike Honeychurch Jan 1 at 22:03