It is said that the logistic function (denoted $y(u)$ below) is derived from the relation:


Does $y(u)=\frac{1}{1+e^{-u}}$ really uniquely satisfy this? I don't see how. It seems to me there must be several functions that can satisfy that. Please help me to see the light.

If I am permitted a follow-up question:

What do I have to do to the first equation so that


satsifies it?


  • 1
    $\begingroup$ You are right, there are infinitely many solutions, though they are all closely related. To determine a unique solution, we need to know an initial condition, like the value of $y(0)$. $\endgroup$ Feb 3, 2012 at 7:20
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    $\begingroup$ I guess you should multiply LHS by $\frac{u}{2}$... $\endgroup$
    – Pedja
    Feb 3, 2012 at 7:32

1 Answer 1


No, $\frac{1}{1 + e^{-u}}$ is not the only function satisfying the equation. So would any $y$ of the form $y(u) = \frac{1}{1 + C e^{-u}}$ for any fixed real number $C$, and so would $y$ given by the formula $y(u) = 0$. (These functions I've listed are it, though, as far as differentiable functions defined on all of $\mathbb{R}$ go; there aren't any others.) I will give a link on how to arrive at this result in a paragraph or two.

The function $\frac{1}{1 + e^{-u}}$ may be the only function satisfying both the differential equation and some initial condition of interest to the person asserting the uniqueness. If you are getting the uniqueness assertion from a professor or textbook, they probably have that in mind, even if they forgot to say it.

There is a general method for solving equations like these, but it is best learned from a textbook. The Wikipedia pages on separable differential equations and separation of variables are a start--- but better would be looking for "separable differential equations" in the index of the nearest calculus or differential equations book. Wikipedia (and the Internet in general, for basic material like this) is more concerned with methods for writing down formulas that solve a differential equation, than proving that the formulas produced by the method are the only solutions to the differential equation.

I'm not sure what your second question precisely is: there's nothing you can "do" to the original equation to make it have that other solution, beyond just changing the equation entirely. So really you're asking, what differential equation might have this other function as a solution.

In general, "given a function, find a differential equation satisfied by it" is not that interesting of a problem (unless you are looking for differential equations within some specific restricted class, or have other more specific goals in mind). Let's see why. If $y(1 + u^{-2}) = 1$, then differentiating both sides with respect to $u$ and using the product rule you find that $y(-2 u^{-3}) + y'(1 + u^{-2}) = 0$. This a differential equation and it is satisfied by the original function $y$ (and other functions too). So that's that.

Maybe this equation is not what you want because it explicitly involves the variable $u$. But one can prove that there is no differential equation of the form $y'(u) = F(y(u))$ satisfied by the function $y(u) = \frac{1}{1 + u^{-2}}$.

To see this, suppose there were.

Since $y(1) = y(-1) = \frac{1}{2}$ (just compute using the formula for $y(u)$) we would conclude by putting in $u = 1$ and $u = -1$ in the equation $y'(u) = F(y(u))$ that $y'(1) = F(\frac{1}{2})$ and that $y'(-1) = F(\frac{1}{2})$ and hence that $y'(1) = y'(-1)$. But you can calculate (use the formula for $y(u)$ to find a formula for $y'(u)$, and then put in $1$ and $-1$) that $y'(1)$ and $y'(-1)$ are not equal. So no such $F$ can exist.


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