Thank You.

It worked. Thank you for your kind reply.

The problem is sorted out.

> eq3 := factor(simplify(Eq3, {2*(2-sigma[t])*gamma*Kn/(sigma[t]*(gamma+1)*Pr) = B}));
                              a = T[w] + H b B
> solve({eq1, eq2, eq3}, {a, b, c});
          /    delta[t] T[w] + 2 T[inf] H B      2 (T[inf] - T[w])  
         { a = ----------------------------, b = -----------------,
          \          delta[t] + 2 H B            delta[t] + 2 H B   

                        T[inf] - T[w]       \
           c = - --------------------------- }
                 delta[t] (delta[t] + 2 H B)/
Using factor and simplify instead of subs and pars works. Thanks.

Omitting the { } in pars does not work here.

> pars := 2*(2-sigma[t])*gamma*Kn/(sigma[t]*(gamma+1)*Pr) = B;
                        2 (2 - sigma[t]) gamma Kn    
                        ------------------------- = B
                         sigma[t] (gamma + 1) Pr     
> eq3 := subs(pars, Eq3);
                             2 (2 - sigma[t]) gamma Kn H b
                  a = T[w] + -----------------------------
                                sigma[t] (gamma + 1) Pr   

Maple is not substituting B in the place of the long expression. Any more suggestions please?

No that way I could have chosen infinity at 10 or lower values also. f ' will go from 0 to 0 through positive values only. But how can I claim then that my solution is infinity independent. Because if I change the value of infinity whole solution changes.

Moreover at 15.6 the value of f ' is sufficiently close to 0 but value of g (0) is about 0.15 which is far greater than 0. g (0) goes to zero at about eta=25.

So  can I get rid of the problem?