 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 givesA 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. ThusFor $0<\nu <1$, eliminate time and integrate:The proportional-infall model of galactic chemical evolution therefore gives, for initially pristine gas,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 $\nu \to 0$ 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.