= Solution
Let $\psi(t)$ be the <star formation rate>, $M$ the remaining gas mass, and $S$ the mass put into stars. With zero returned bulk gas, pristine <galactic gas inflow>, and no <galactic outflow>, <mass conservation> gives
$$
\dot S=\psi,\qquad \dot M=-(1-\nu)\psi.
$$
A well-mixed gas reservoir loses existing metals at rate $Z\psi$ and gains newly synthesized metals at rate $y\psi$; the incoming gas brings none. Thus
$$
\frac{d(MZ)}{dt}=y\psi-Z\psi,\qquad
M\dot Z=(y-\nu Z)\psi.
$$
For $0<\nu<1$, eliminate time and integrate:
$$
\frac{dZ}{y-\nu Z}=-\frac{dM}{(1-\nu)M},\qquad
\frac{y-\nu Z}{y-\nu Z_0}=\left(\frac M{M_0}\right)^{\nu/(1-\nu)}.
$$
The <proportional-infall model of galactic chemical evolution> therefore gives, for initially pristine gas,
$$
\boxed{Z=\frac y\nu\left[1-\left(\frac M{M_0}\right)^{\nu/(1-\nu)}\right]}.
$$
\b[The displayed form requires $Z_0=0$]; more generally $Z=y/\nu+(Z_0-y/\nu)(M/M_0)^{\nu/(1-\nu)}$. Also $S=(M_0-M)/(1-\nu)$ for an initially star-free reservoir, so $M/M_0$ is not the <gas fraction of a galaxy> $M/(M+S)$. As a check, the $\nu\to0$ limit is the <closed-box model of galactic chemical evolution> relation $Z=y\log(M_0/M)$; at $\nu=1$ the gas mass is constant and one must instead solve $\dot Z=(y-Z)\psi/M_0$. Retaining a nonzero <stellar yield> while setting bulk mass return to zero is the prescribed trace-metal approximation, not a literal claim that stellar ejecta have no mass.
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