... equation:xh(tn+1)=m−1k=0x(k)0tkn+1k!+hq(q +2)f (tn+1, xph(tn+1)) +hq(q +2)nj=0aj,n+1f (tj, xn(tj)) (1 2)whereaj,n+1=⎧⎨⎩nq+1− (n − q)(n +1)q; j =0(n − j +2)q+1+(n − j)q+1− 2(n − j +1)q+1;1≤ ... transformation:u(x)=v(x) − 0.2 + x3 (1 0 − x1) (3 0)Applying this control law to (2 9) yields:⎛⎝C0Dq1tx1(t )C0Dq2tx2(t )C0Dq3tx3(t )⎞⎠=⎛⎝−(x2+ x3)x1+0.63x2v(x)⎞⎠ (3 1)Let’ ... Txk (0 ) = x(k)0, k =0,1,··· , m − 1 (1 0)This equation is equivalent to the Volterra integral equation given by [35]:x(k)=m−1k=0x(k)0tkk!+1(q)t0(t − s)q−1f (s, x(s))ds (1 1)Consider...