= Quadratic escape desingularization
For $\dot u=u(\mu+u-v)$, $\dot v=\lambda u^2$ in $u>0$, the time change $d\tau/dt=u$ gives the affine linear system $u_\tau=\mu+u-v$, $v_\tau=\lambda u$. Shifting $w=v-\mu$ removes the forcing. Its <eigenvalues> are $(1\pm\sqrt{1-4\lambda})/2$. For $\lambda>1/4$, putting $\omega=\sqrt{4\lambda-1}/2$ gives
$$
u(\tau)=e^{\tau/2}\left[h\cos(\omega\tau)+\frac{\mu+h/2-v_0}{\omega}\sin(\omega\tau)\right].
$$
For fixed $\mu>0$ and initial $h,v_0$ tending to zero, the first return to small positive $u$ occurs after approximately $\pi/\omega=2\pi/\sqrt{4\lambda-1}$. This is a half-turn, not a full turn. At $\lambda=1/4$, $u=e^{\tau/2}[h+(\mu+h/2-v_0)\tau]$ does not return and grows without bound. The loss of return diagnoses escape to infinity rather than a finite-amplitude <homoclinic orbit>. Physical time obeys $dt=d\tau/u$ and need not share the divergent desingularized travel time.
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