Let $\lambda >0$. Consider the non-linear system $$ \begin{cases}\dot{x}=x^2\\\dot{y}=-\lambda y\end{cases} $$
The aim is to get an idea of the phase portrait.
There are some things I can see directly:
- First of all, $(0,0)$ is an equilibrium.
- On the $x$-axis, we are moving to the right.
- On the positive $y$-axis we are moving down towards the equilibrium $(0,0)$.
- On the negative $y$-axis we are moving up towards the equilibrium $(0,0)$.
But what happens if we start with an initial pair $x(0)=x_0, y(0)=y_0$ in the four quadrants?
In order to get an idea, I determined the solutions:
$$ x(t)=\frac{1}{\frac{1}{x_0}-t},\qquad y(t)=y_0\cdot\exp(-\lambda t). $$
Concerning the solution $y$, if $y_0<0$, $y$ grows exponentially, if $y_0>0$, it decays exponentially.