$x'=x^2$ unstable solutions when $x(0)\geq 0$ but asymptotically stable when $x(0)\leq0$

Question

How do I show that the differential equation $x'=x^2$ has unstable solutions when $x(0)\geq 0$ but asymptotically stable solutions when $x(0)\leq0$?

Usually, I look at the eigenvalues of the matrix to determine stability of solutions, but since there is no matrix here, how do I approach this?

Edit: the solutions are $x(t)=\frac{1}{c-t}$, but what can I do with this?

Answer 1

suppose $x(0) = k.$ then the solution is $$x = \frac 1{t + 1/k }, -1/k < t < \infty$$ the graph is monotone increasing on the domain and $\lim_{t \to \infty} x(t) = 0.$

this solution approaches zero but not expentailly; it approaches zero like $\frac1t.$ i don't this the solution is asymptotically stable; just stable.

Answer 2

Qualitative analysis of ODEs gives an answer to the questions. Namely, the behaviour of solutions of $x' = f(x)$ follows from the form of function $f(x)$. Zeros of $f(x)$ gives stationary points, so $x$ approaches asymptotically to them, provided that an appropriate initial condition is taken. From the equation, it follows that $f(x)$ is a rate of variation of $x$. Therefore, if $f(x) > 0$, then x grows, while, when $f(x) < 0$, $x$ decreases. It is helpful to plot $f(x)$ as a function of $x$, then those $x$ that have positive (negative) $f(x)$ are increased (are decreased). These arguments can be easily appiled for the equation in the question.