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Given the ODE $\dot{x} = f(t,x)$, $f(t,0)=0$, we know that $x\equiv 0$ is a solution. This solution is Lyapunov stable if for any $\epsilon>0$, $t_0\ge 0$, there exists $\delta>0$ such that $|x_0|<\delta$ implies $|x(t,t_0,x_0)|<\epsilon$ for $t\in [t_0,\infty)$.
To be clear, $x(t,t_0,x_0)$ is the unique solution to $\dot{x} = f(t,x)$ such that $x(t_0,t_0,x_0) = x_0$.

In particular, I'm asked to find out the stability of every solution to $\dot{x} = -x(1-x)$, not just the $0$ one. I'm just a bit unsure of how to approach the problem. I've solved the ODE and obtained $$ x(t) = \frac{C}{Ce^t+1}.$$ Solving for $C$, we obtain $C = \frac{x_0}{1-x_0}$. I suppose what my question is, from here, in order to show stability, I just need to find delta such that $|x_0|<\delta$ implies the general solution above is bounded by $\epsilon$, correct? (Or show that it can't be stable). But I'm a bit confused on how to show stability for "every" solution, i.e., the solution $x(t)$ I found above.

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    You want to show stability of solutions or of equilibrium points? For example Wikipedia says Lyapunov stability is defined as stability of eq. points, and I've never heard of stability of solutions. Maybe your problem is to determine all equilibrium points and whether or not they're stable. – 2017-02-13
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    @mathematician The book I'm using says "solutions" instead of equilibrium points, and gives a brief mention that we can discuss it for any solution and not just the 0 one. The wording isn't very clear (to me) unfortunately. But in any case, the equilibrium points are indeed solutions anyway, so maybe it was meant to discuss their stability – 2017-02-13

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I hope I am understanding the question correctly (otherwise feel free to criticize me). I'd formulate an answer as follows:

$$x'=-x(1-x)$$

has the equilibria (or 'solutions') $x_0=0$ and $x_1=1$ (since for these values we have $x'=0$). Actually you don't need to compete the solution of the ODE itself. Here we have beside the $0$ solution also $x_1$ as a fixed point.

  • for $x<0$ we have $$x'=\underbrace{-x}_{>0}(\underbrace{1-x}_{>0})>0$$
  • for $00})<0$$
  • for $x>1$ we have $$x'=\underbrace{-x}_{<0}(\underbrace{1-x}_{<0})>0$$

We conclude:

  • the trajectories on the left side from $x_0$ move towards $x_0$
  • the trajectories between $x_0$ and $x_1$ move towards $x_0$
  • the trajectories on the right side from $x_1$ move towards $\infty$

Therefore $x_0$ is an asymptotically stable solution/equilibrium and $x_1$ is unstable in the sense of Lyapunov.

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    Thank you, I actually think that's exactly what it was looking for. Since I'm also supposed to investigate asymptotic stability, I think the general solution would help, since sending $t\to \infty$ would give us that $x(t)\ to 0$, whence asymptotic stability (though I'm sure there is some other analysis one can do as we cannot always solve the ODE) – 2017-02-14
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    @Curious You are right, there are clearly cases where the solution could help to determine the stability. But here it is pretty easy. Since all trajectories move towards $x_0=0$ we automatically conclude that $x_0$ is asymptotically stable. It could help you to draw a phase portrait. Later one you get some other nice tools to determine stability. For example looking at the linearization of the ODE around one equilibrium. – 2017-02-14