Differential equation which has following solution $y=\frac{1}{1+\exp(ax)}$

Question

Is there any linear differential equation which has following solution $$y=\frac{1}{1+\exp(ax)}$$ $a$ is constant. something like: $$ y'' + by' +cy + \alpha = 0$$ where $b$, $\alpha$ and $c$ are constants. When I take derivative it looks more and more complicated.

---

Update: If not possible, What if the coefficients ($b$ and $c$) can be like: $$ b = b_0 y, \; c = c_0 y + d_0 $$

Answer 1

Note that $z:=y^{-1}-1=e^{a x}$ satisfies $z''=az'=a^2 z$. From this we observe that $$y'=-z'(1+z)^{-2}=-a z y^2=-a(y-y^2)\implies y'+a(1-y)y=0.$$ This is a nonlinear first-order ODE, which is separable and so can be integrated to obtain $y(x)=(1+C e^{a x})^{-1}.$ For $C=1$ this reproduces the given solution, but note that this solution is nonlinear in $C$.

From this nonlinear ODE we can create another: $$y''=\dfrac{d}{dx}\left[-a (y-y^2)\right]=-a (1-2y)y'\\\implies y''+a(1-2y)y'=0\\ \implies y''-2ayy'-a^2 (1-y)y=0$$

So now we have a second-order nonlinear ODE, but whose coefficients are linear in $y$.

Answer 2

Not a constant-coefficient homogeneous linear differential equation. The Laplace transform of the solution of a constant-coefficient de is a rational function. This function's Laplace transform is not rational (it has infinitely many poles).

Answer 3

You want the logistic equation: wikipedia. mathworld, William Stein's writeup. Caveat: this is a nonlinear equation.

Answer 4

Only for completeness (and for playing around).

You can show that the following (false) 2nd order linear ODE:

$$ y'' + a \tanh{\frac{a x}{2} } \, y' = 0, \quad y=y(x),$$ has $1/(1+e^{ax})$ as a solution (the other one is $1/(1+e^{-ax})$!).

Hint: you may want to work out on the definition of $\tanh$, particularly in the case of double/half angles.