Vtgyjfy

I encountered the following ODE:

$$frac{dx}{dt} = x(1-x)$$

Of course, this can easily be solved with separation of variables:

$$implies int frac{dx}{x(1-x)} = int dt implies ln bigg|frac{x}{1-x} bigg| = t+C$$

This solution is valid on $Bbb R backslash {0,1}$, so there would be no problem with imposing initial condition $x(0) = x_0$ as long as $x_0 neq 0,1$.

But, my question is how would we solve this ODE if the initial condition was indeed $x_0=0$ or $x_0=1$?

Intuitively, I see that these are fixed points so that $x equiv 0$ or $x equiv 1$ in these cases. But I guess I would like to know whether there is a way to express the solution of the ODE given general initial condition?

Think about the initial condition as a parameter. Since
$$
ln left| frac{x}{1-x}right| = e^t + C
$$
we know
$$
frac{x}{1-x} = A e^t
$$
where $A = e^C$. Evaluating at $t=0$ gives
$$
A = frac{x(0)}{1-x(0)}
$$
and so
$$
x(t) = frac{frac{x(0)}{1-x(0)} e^t}{1 + frac{x(0)}{1-x(0)} e^t}
= frac{x(0) e^t}{1+x(0)(e^t-1)}
$$
So now think about the function $gamma(s,t)$, where for each $s$, $t mapsto gamma(s,t)$ is the solution to $x' = x(1-x)$, $x(0) = s$. Using the above, we see that
$$
gamma(s,t) = frac{se^t}{1+s(e^t-1)}
$$
Notice $lim_{sto 0} gamma(s,t) = 0$ and $lim_{sto 1} gamma(s,t) = 1$ for all $t$. So the two equilibrium cases are the limits of the non-equilibrium cases as the initial conditions tend toward the critical points.

If you intially start with $x_0 = 0$ or $x_0 = 1$, then there won't be a change in $x$ over time because in both cases you have:
$$frac{dx}{dt} = 0$$

So your $x$ won't change and will remain the same value so in this case you have $x = 0$ and $x=1$. You can't solve the ODE generally and analysing $x_0 = 0$ or $x_0 = 1$ at the end with your general solution. You need to consider them seperately. There isn't also a way to write the solution of the ODE in one form including $x = 0$ and $x=1$. You need to write the solution depending on which begin condition you use:
$$x = 0 hspace{3mm} text{if} hspace{3mm} x_0 = 0\ lnleft|frac{x}{1-x} right| = t+C hspace{3mm} text{if} hspace{3mm} x_0 = mathbb{R}backslash{0,1}\x = 1 hspace{3mm} text{if} hspace{3mm} x_0 = 1\$$

Solving for $x$ gives

$$ x(t) = frac{Ae^t}{1+Ae^t} $$

where $A = e^C$

Note that $x(0) = dfrac{A}{1+A}$

The initial condition $x(0)=0$ occurs when $A=0$. This gives the constant solution, as predicted.

The initial condition $x(0)=1$ occurs in the limit $Atoinfty$. This also gives the constant solution.

I got $$x(t)=frac{e^t}{e^t+e^{C}}$$ then we get $$x(0)=frac{e^{0}}{e^{0}+e^{C}}$$

If $x_0 = 0$, then the function $x(t)=0$ satisfies both the initial condition and the differential equation.

Similarly, if $x_0=1$, the function $x(t)=1$ does the same.

First of all, solve for $x$ in the cases $x_0 neq -1, 0$:

$ln left| frac{x(t)}{1-x(t)} right| = t + C$,
hence
$x(t) = frac{e^{t+C}}{1+e^{t+C}}$.
As $C = ln left| frac{x_0}{1-x_0} right|$ we have
$$x(t)
= frac{e^{t+ln left| frac{x_0}{1-x_0} right|}}{1+e^{t+ln left| frac{x_0}{1-x_0} right|}}
= frac{left| frac{x_0}{1-x_0} right|e^t}{1+left| frac{x_0}{1-x_0} right|e^t}
= frac{|x_0|e^t}{|1-x_0|+|x_0| e^t}.$$

You asked for a closed formula for all $x_0$. Well, this formula is also valid for $x_0 = 0$ or $x_0=1$.