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Working through some past papers for my Differential Equations class, I've found difficulty in this problem:

Problem

Considering the plane autonomous system

$$\frac{dx}{dt}=x+2y, \quad \frac{dy}{dt}=-2x-y-\frac{1}{2}x^3$$

I'm looking to find and classify the critical points and show that the phase plane paths are

$$x^2+xy+y^2+\frac{1}{8}x^4=\text{constant}$$

Progress

I've established that the only critical point on the plane is at $(0,0)$, and then considered roots of the characteristic equation of the matrix

$$ \left( \begin{array}{ccc} X_x & X_y \\ Y_x & Y_y \end{array} \right)$$ where $X=\frac{dx}{dt}$ and $Y=\frac{dy}{dt}$ evaluated at $(0,0)$ which were $\lambda = \pm \sqrt3 i$; I think this leads us to classify the critical point as a centre though I'm not sure.

I can't see how to establish the equations of the critical points. Any help would be greatly appreciated. Regards as always, MM.

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Hint: differentiate $x^2 + x y + y^2 + x^4/8$ with respect to $t$, getting $dx/dt$ and $dy/dt$ from the differential equations. – Robert Israel Jan 12 '12 at 21:13
@RobertIsrael: Differentiating yields $\frac{dx}{dt}(2x+\frac{1}{2}x^3+y) + \frac{dy}{dt}(x+2y)=0$, which looks promising. Not sure how this gives us the phase planes though. – Mathmo Jan 12 '12 at 21:19

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