First integral of an ordinary differential equation ($I_t+\nabla I\cdot f=0$)

A first integral of $y^{\prime }=f(t,y)$ is a differentiable function $I(t,y)$ whose value is constant along solutions through every time and state in its domain. The chain rule makes this equivalent to the displayed pointwise identity. For $I(y)=y^{T}Sy$ with symmetric $S$, it becomes $y^{T}Sf(t,y)=0$. This identity at the implicit stage is what proves exact Runge-Kutta conservation of quadratic invariants for the implicit midpoint rule. Invariance asserted only on the image of one initial-time flow needs that image to contain the stage under consideration.

For this cubic vector field, every non-axis trajectory has the first integral $I(x,y)=1/y-y/x^2$. Direct differentiation gives $\nabla I\cdot (\dot{x},\dot{y})=0$ where $xy\ne 0$. Equivalently the trajectories satisfy $x^2-y^2=C{x}^{2}y$. The axes are separate invariant trajectories, and the diagonals are the cases $C=0$. For each fixed finite $C$, approach to the origin forces $y^2/x^2=1-Cy\to 1$, so non-axis trajectories have diagonal limiting directions.

The origin is repelling in forward time. For $V=x^2+y^2$, $\dot{V}=2x^4+2x^2{y}^2+4y^4\ge V^2$, so every nonzero trajectory eventually exits any small disk. The zero Jacobian matrix at the origin leaves the linearized system constant, demonstrating why nonlinear terms are needed for this stability classification.

For $\dot{x}=y^3-y$, $\dot{y}=x-y^2$, set $a_{\pm }=(1\pm \sqrt{3})/2$ and $F_{\pm }=x-1-a_{\pm }(y^2-1)$. Direct differentiation gives ${\dot{F}}_{\pm }=-2a_{\pm }y{F}_{\pm }$, proving the displayed first integral. The invariant parabolas meet at saddles $(1,\pm 1)$ and bound a heteroclinic cycle. Inside that lens, the logarithm of $I$ has a strict nondegenerate maximum at the origin, proving a nonlinear center equilibrium, rather than inferring one solely from imaginary Jacobian eigenvalues.