## 5 Reputation

13 years, 305 days

Thank You.

## Syntax...

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.

## Syntax...

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?

## infinity=15.6...

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?

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