Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 2 / 8C (Differential Equations)

Along a trajectory, the chain rule givesThus $\boxed{H=x^2+y^2-y^4}$ is a first integral, and every trajectory lies on one of its level sets.

A critical point of a dynamical system satisfies $x=0$ and $y-2y^3=0$, so the critical points areThe Jacobian matrix of the vector field isAt the origin its eigenvalues are $\pm i$. Moreover, $H=x^2+y^2+O(y^4)$ has a strict local minimum there, so the origin is a stable center of a planar dynamical system. At either other critical point,has eigenvalues $\pm \sqrt{2}$, so both points are unstable saddle points. Their common value $H=1/4$ identifies the special separatrices shown below.

The initial point has $H=1/4$. On the branch leaving $(1/2,0)$ toward negative $y$,Since $\dot{y}=-x$, the required scalar equation isIts solution isAt $(x,y)=(1/4,-1/2)$, this requires $\tanh (t/\sqrt{2})=1/\sqrt{2}$. Using the inverse hyperbolic tangent,