(*  Fourth-order RK  Equations  ---  Solution *)
(*  Before running, just set a[1]=yourValue and a[2]=anotherValue  *)
(*  Do not use these values, but for example, a[1]= 3/7;  a[2] = 5/7; *)

Clear["Global;`*"];
a[1]= 3/7;  
a[2] = 5/7;
soln={c[1]->((-1+2 a[2])/(12 (-1+a[1]) a[1] (a[1]-a[2]))),
b[3,1]->(((-1+a[1]) (-2+a[1]+(5-4 a[2]) a[2]))/(2 a[1] (a[1]-a[2]) (3-4 a[2]+a[1] (-4+6 a[2])))),
b[2,1]->(((a[1]-a[2]) a[2])/(a[1] (-2+4 a[1]))),
b[3,2]->(((-1+a[1]) (-1+2 a[1]) (-1+a[2]))/((a[1]-a[2]) a[2] (3-4 a[2]+a[1] (-4+6 a[2])))),
c[2]->((1-2 a[1])/(12 (a[1]-a[2]) (-1+a[2]) a[2])),
a[3]->1,
c[3]->((3-4 a[2]+a[1] (-4+6 a[2]))/(12 (-1+a[1]) (-1+a[2])))}

(*  By the way, "/.soln" means that, in the given expression, you are substituting the values for the parameters given in "soln" *)

c[0]=1-c[1]-c[2]-c[3]/.soln   (* since the sum of the c's has to equal 1, we get for c[0]: *)

(*  Now,since the sum of the b[k,j]'s has to equal a[k]
for k=1,2,and 3, we also get that...*)
b[1,0]=a[1]/.soln

b[2,0]=a[2]-b[2,1]/.soln

b[3,0]=a[3]-b[3,1]-b[3,2]/.soln