we consider $x' = -x^3,~ y'=-x^2 y$. I would like to show that for positive initial values $x(0)=x_0>0,~ y(0)=y_0>0$ solutions tend to the origin.
Now I know that this is explicitly solvable but I would like to prove this without using the analytical solution (as this is just a toy model for a more complicated problem I have).
The equilibrium $(0,0)$ is not asymptotically stable because the whole axis $\{0\}\times \mathbb R$ is a set of fixed points. So I can't really use Lyapunov theory: If we take for example $V(x,y) = \frac1{2}(x^2+y^2)$, we get
$\dot V(x,y) = \langle \nabla V(x,y),(x',y')\rangle = -x^2 \cdot (x^2+y^2)$, which is flat along the y-axis.
This is to be expected, of course. But I would like to show that at least for a subset of initial values (with the only restriction in this case being $x_0\neq 0$) we have convergence.
From $x'=-x^3$ we obtain: $$-\dfrac{2}{x^3}x'=2\to \dfrac{1}{x^2}=2t+C_1$$letting $t=0$ we get $C_1=\dfrac{1}{x_0^2}$ therefore$$x^2=\dfrac{1}{2}\dfrac{1}{t+\dfrac{1}{2x_0^2}}$$by substituting in $y'=-x^2y$ we get:$$\dfrac{y'}{y}=-x^2=-\dfrac{1}{2}\dfrac{1}{t+\dfrac{1}{2x_0^2}}$$integrating leads us to:$$\ln y+C_2=-\dfrac{1}{2}\ln{t+\dfrac{1}{2x_0^2}}$$or $$y=C_3(t+\dfrac{1}{2x_0^2})^{-\dfrac{1}{2}},\qquad\qquad\qquad C_3>0$$by substituting initial condition we finally have:$$y=\dfrac{y_0}{\sqrt{1+\dfrac{2}{x_0^2}t}}$$and $$x=\sqrt{\dfrac{1}{2}\dfrac{1}{t+\dfrac{1}{2x_0^2}}}$$by $t\to\infty$ both $x$ and $y$ get to zero.