There is a hint in the error message that the limit cannot be computed. When you have symbolic coefficients, it is possible, even likely, that the existence of the limit depends on them. If we specify the coefficient of t'[e]
is positive, we get a result:
Assuming[Pr (1 - Exp[-K η])/η > 0,
DSolve[t''[e] + Pr (1 - Exp[-K η])/η t'[e] == 0 &&
t[0] == 1 && t[Infinity] == 0, t, e]
]
(* {{t -> Function[{e}, E^((e (-1 + E^(-K η)) Pr)/η)]}} *)
If it is not positive, then there is no solution:
Assuming[Pr (1 - Exp[-K η])/η < 0,
DSolve[t''[e] + Pr (1 - Exp[-K η])/η t'[e] == 0 &&
t[0] == 1 && t[Infinity] == 0, t, e]
]
(* {} *)
Likewise for the boundary case, if the coefficient is zero, which is probably not the case anyway.
Porbably one knows something about the coefficients from the context of the problem in which it arises. If the critical value of the coefficient is not obvious (greater/less than zero in this case), one might inspect the general solution. My eyes are weak so I used a big variable:
t[XXXXXX] /. DSolve[t''[e] + Pr (1 - Exp[-K η])/η t'[e] == 0 && t[0] == 1, t, e] // Simplify
(*
{((-1 + E^(K η)) Pr - E^(K η) (-1 +
E^(((-1 + E^(-K η)) Pr XXXXXX)/η)) η C[1])/((-1 + E^(K η)) Pr)}
*)
The variable XXXXXX
appears only in the exponential so the sign of its coefficient will determine whether the limit exists or not. Vitaliy's advice to replace a complicated coefficient with a simple symbol is good, too.
Pr (1 - Exp[-K η])/η
toPr (1 - Exp[-Ke])/K
, so it is still a linear DE with constant coefficients. That does not change the validity of the approaches in Vitaly's and my answers. All it does is put our constants out of step with the question's. $\endgroup$Pr (1 - Exp[-Ke])/K
equal to a since it includes 'e' now which is an independent variable? I wonder how will it give solution in form of required function. $\endgroup$Pr (1 - Exp[-K e])/K
, with a space between theK
and thee
.Ke
would be interpreted as a single symbol. $\endgroup$DSolve
uses the symbolK
(and the symbolC
) in solving equations. You should probably avoid usingK
as a variable. Try executing? K
. $\endgroup$