Bob, what kinds of quantities are k_1 & k_2 supposed to be? Since the equation is trancendental we don't expect to find a general solution formula. But we can make some general observations about the solutions when k_1 & k_2 are real-valued.
If k_1 & k_2 are supposed to be real numbers then the equation has no real solution for x when k_1*k_2 > 1/e.
If k_1*k_2 = 1/e then x = 1/k_2.
If 0 < k_1*k_2 < 1/e then there are *two* real solutions for x of the same algebraic sign as k_2 (one larger than 1/k_2 and the other one between 0 & 1/k_2).
If k_1 = 0 then x = 0.
If k_1*k_2 < 0 then there is one real solution for x whose algebraic sign is opposite of k_2.
Suppose we define the real-valued function f(u) of the real argument u according to:
f(u) = = u*exp(-u). Then in general for real k_1 & k_2 the value(s) of the solution x can be formaly written as:
x = (1/k_2)*f^-1(k_1*k_2) where f^-1(y) means the inverse function of f, i.e. y = f( f^-1(y)).
The function f^-1(y) is single-valued and negative for a negative argument. It is zero when its argument u is zero. It is double-valued with both values being positive for positive arguments that are also less than 1/e (in which case one value is between zero & one and the other value is greater than 1). It has the value 1 when its argument is exactly 1/e. And it has no real values at all for arguments larger than 1/e.
David Bowman
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From: phys-l-bounces@carnot.physics.buffalo.edu [phys-l-bounces@carnot.physics.buffalo.edu] On Behalf Of Spinozalens@aol.com [Spinozalens@aol.com]
Sent: Thursday, May 13, 2010 2:33 PM
To: phys-l@carnot.physics.buffalo.edu
Subject: [Phys-l] Can Anyone Find a Solution For This Equation?
Can anyone find a general solution for the equation below for the variable
x