Consider the differential equation
$$x''+(2a)x'-(b^2/2)x+x^3=0 \tag{1}$$
where $a,b >0$ are constants.
(i) Write differential equation (1) as a first order dynamical system.
(ii) Determine the equilibrium point(s) of the system found in part (i).
(iii) Classify the equilibrium point $$(b/\sqrt2,0).$$
For (i) I got $$x'=y$$ $$y'=-x^3+(b^2/2)x-(2a)y$$
For (ii) I got equilibrium points $$(0,0), (b/\sqrt2,0), (-b/\sqrt2,0)$$
Now for (iii) I moved the eq point $(b/\sqrt2,0)$ to the origin. So I let $X=x-b/\sqrt2$ and $Y=y$ and then found the associated linear system as $$X'=Y$$ $$Y'=(b^2)X-(2a),Y$$ and then found the eigenvalues $$\lambda=-a+\sqrt{a^2+b^2}$$ and $$\lambda=-a-\sqrt{a^2+b^2}$$ and found that the first one is $>0$ and second one is $<0$. This means that it is a saddle point right? Do we have to find anything else to determine that it is a saddle point? And more importantly, i was told that we need to think about the linearization theorem to finish the question but i don't know how. Please help and if i have done any mistakes, please let me know!!