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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?

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  • $\begingroup$ Its perhaps important to note that this equation satisfies the hypothesis of Picard-Lindelof so the solution is unique. $\endgroup$ – Calvin Khor Jan 22 at 14:28
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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 $\lim_{s\to 0} \gamma(s,t) = 0$ and $\lim_{s\to 1} \gamma(s,t) = 1$ 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.

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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: $$x = 0 \hspace{3mm} \text{if} \hspace{3mm} x_0 = 0\\ \ln\left|\frac{x}{1-x} \right| = t+C \hspace{3mm} \text{if} \hspace{3mm} x_0 = \mathbb{R}\backslash\{0,1\}\\x = 1 \hspace{3mm} \text{if} \hspace{3mm} x_0 = 1\\$$

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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 $A\to\infty$. This also gives the constant solution.

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

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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.

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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 $x_0=1$.

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