= Solution
Use <dimensionless variables> $X=X_*x$, $Y=Y_*y$, $Z=Z_*z$, $t=T\tau$, with
$$
\boxed{Y_*=\frac1b,\quad Z_*=\frac g{mb},\quad X_*=\frac{lg}{rmb},\quad T=\frac{lg}{rmk}.}
$$
Substitution gives the requested system, with dots now meaning differentiation with respect to $\tau$, and
$$
\boxed{\alpha=\frac{ab}k,\qquad\epsilon=\frac{rm^2k}{l^2g^2},\qquad\delta=\frac{rk}{lg},\qquad\gamma=\frac{3c}{b^2}.}
$$
In particular the same scale for $Y$ makes the first two coefficients of the cubic equal in magnitude.
Its <derivative> is $f'(y)=1-2y+\gamma y^2$. For $0<\gamma<1$ its two simple zeros are
$$
\boxed{y_\pm=\frac{1\pm\sqrt{1-\gamma}}\gamma.}
$$
The cubic increases below $y_-$, decreases between the two <fold points of a critical manifold>, and increases above $y_+$. It has a <local maximum> at $y_-$ and a local minimum at $y_+$.
After a rapidly decaying <transient response>, $z\simeq f(y)$ when its relaxation is faster than the other variables. The reduced <autonomous system> is
$$
\boxed{\dot x=\alpha-y,\qquad\dot y=\frac{x-f(y)}\epsilon.}
$$
For $\epsilon\ll1$, the fast $y$ dynamics at fixed $x$ attracts to the branches where $f'(y)>0$ and repels from the middle branch. On the low branch $y<y_-<\alpha$, $x$ increases slowly. At its <fold point of a critical manifold> $x=f(y_-)$, the attracting low branch disappears and $y$ jumps upward, with nearly constant $x$, to the high branch. On that branch $y>y_+>\alpha$, $x$ decreases slowly. At $x=f(y_+)$ it jumps back to the low branch. These slow arcs and fast jumps form the singular relaxation cycle. For small positive $\epsilon$, and parameters bounded away from the <fold point of a critical manifold>-endpoint cases, the nearby attracting periodic motion has the same pattern. The <equilibrium> $(f(\alpha),\alpha)$ is on the repelling branch: its <Jacobian matrix> has <determinant> $1/\epsilon>0$ and <trace> $-f'(\alpha)/\epsilon>0$.
The limiting landing values are $3/\gamma-2y_-$ after the upward jump and $3/\gamma-2y_+$ after the downward jump, obtained from the third root of $f(y)=f(y_\pm)$. Thus the two vertical segments in the <phase plane> plot really connect roots of the same fast equation.
\Image[/past-exam-of-the-differential-equations-and-applications-course-of-the-university-of-oxford/2007-relaxation-cycle.png]
{title=Cubic folds and the singular relaxation cycle: slow drift on attracting branches and fast jumps at nearly fixed x}
If $\gamma>1$, $f'$ is strictly positive everywhere, so the cubic is <monotone> and has no <fold points of a critical manifold>. The reduced <equilibrium> is asymptotically stable. In fact
$$
V=\tfrac12[x-f(\alpha)]^2+\tfrac\epsilon2(y-\alpha)^2,\qquad
\dot V=-(y-\alpha)[f(y)-f(\alpha)]\leq0.
$$
The only invariant trajectory in the zero-<derivative> set is the <equilibrium>, so bounded reduced solutions converge to it; <relaxation oscillations> disappear.
The stronger condition $\boxed{\delta\ll\epsilon}$ matters during a fast jump: $y$ then varies on the $\epsilon$ time scale, not on the slow unit scale. Only if $z$ adjusts faster still may one keep $z\simeq f(y)$ through that jump. Merely imposing $\delta\ll1$ would not justify this reduction on all pieces of the cycle. This is a <three-time-scale cubic relaxation oscillator>.
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