sim_difeq.c solving 4 equations 
simple numeric check, answer 1, 2, 3, 4 
X= 0.100000000000000 Y[0]= 0.100000000000000 Y[1]= 0.200000000000000 
                     Y[2]= 0.300000000000000 Y[3]= 0.400000000000000 
X= 0.200000000000000 Y[0]= 0.200000000000000 Y[1]= 0.400000000000000 
                     Y[2]= 0.600000000000000 Y[3]= 0.800000000000000 
X= 0.300000000000000 Y[0]= 0.300000000000000 Y[1]= 0.600000000000000 
                     Y[2]= 0.900000000000000 Y[3]= 1.200000000000000 
X= 0.400000000000000 Y[0]= 0.400000000000000 Y[1]= 0.800000000000000 
                     Y[2]= 1.200000000000000 Y[3]= 1.600000000000000 
X= 0.500000000000000 Y[0]= 0.500000000000000 Y[1]= 1.000000000000000 
                     Y[2]= 1.500000000000000 Y[3]= 2.000000000000000 
X= 0.600000000000000 Y[0]= 0.600000000000000 Y[1]= 1.200000000000000 
                     Y[2]= 1.800000000000000 Y[3]= 2.400000000000000 
X= 0.700000000000000 Y[0]= 0.700000000000000 Y[1]= 1.400000000000000 
                     Y[2]= 2.100000000000000 Y[3]= 2.800000000000001 
X= 0.800000000000000 Y[0]= 0.800000000000000 Y[1]= 1.600000000000001 
                     Y[2]= 2.399999999999999 Y[3]= 3.200000000000001 
X= 0.900000000000000 Y[0]= 0.900000000000000 Y[1]= 1.800000000000001 
                     Y[2]= 2.699999999999999 Y[3]= 3.600000000000001 
X= 1.000000000000000 Y[0]= 1.000000000000000 Y[1]= 2.000000000000001 
                     Y[2]= 2.999999999999999 Y[3]= 4.000000000000002 

fourth order check 1, 1/2, 1/6, 1/24 
X= 1.000000000000001 Y[0]= 1.000000000000001  Y[1]= 0.500000000000001 
                     Y[2]= 0.166666666666667  Y[3]= 0.041666666666667 
Y[3]*24.0 =  1.000000000000001 

sim_difeq.c finished