I have been given the system: $$\begin{cases}\dot{x}=-x+y^2\\\\ \dot{y}=x^2-y\end{cases}$$ (Note: the left hand sides should be x and y with a dot on top however I can't quite find how to write that, if anyone could edit that in for me I would appreciate it)
From here I've been tasked to find the equilibria and determine their stabilities.
I'm happy to show working if necessary however I found: Equilibria are $(0,0)$ and $(1,1)$.
From here: $(0,0)$ is a first kind asymptotically stable node as for $Df(0,0)$, the eigenvalues $λ_{1,2} =-1<0$ with $λ_1$=$λ_2$. Does this count as 'determining the stability of the equilibria by linearisation', I'm unsure exactly what this means.
I am however having trouble with $(1,1)$: I can see that $Df(1,1)$ has negative trace (-2) and determinant (-3), with eigenvalues $λ_{1,2}=1 \pm 2$. Looking through my notes I'm unsure how I can use this information to get the result I want. Am I approaching this the right way? If so, how can I use what I've found to determine $(1,1)$'s stability (and why if you can)?
Any pointers provided would be greatly appreciated, I'm trying to get ahead of my final weeks of content so I don't feel rushed with exam szn looming, so I'm missing a lot of understanding which I'm aiming to improve by getting ahead.
Everything you obtained looks OK to me. Except I obtain $\{ \lambda_1, \lambda_2 \} = \{ -3, 1 \}$ at the equilibrium $(1,1)$ (not $\{ -1, 3 \}$). So you have a positive eigenvalue. Let us look at what happens when you consider a small perturbation near this equilibrium. Write $(x,y) = (1+a,1+b)$. By linearization, these small perturbations evolve following the system $$ \dot{a} = -a + 2b \\ \dot{b} = 2a - b $$ In short $\dot{v} = M v$ where $v = (a,b)$ and $M$ is the matrix. The fact that you found a positive eigenvalue means that there exists an initial data $v_0 = (a_0,b_0)$ (the eigenvector associated with the eigenvalue) such that $e^{3t}v_0$ is a solution to this equation. The norm of this solution grows with time. Even if you started with the initial data $(1,1) + \varepsilon v_0$, which is very close to the equilibrium, the solution would (at first order) be $(1,1) +\varepsilon e^{3t} v_0$, growing further and further away. So this equilibrium is unstable.