= Solution
During ejection the <mitral valve> is closed, so $Q_v=0$. Adding the arterial and ventricular balances cancels aortic flow:
$$
R_cC_a\dot p_a+R_c\frac{d}{dt}(p_{LV}/E_{LV})=-(p_a-p_v)\simeq-p_a,
$$
where the last approximation assumes $p_v\ll p_a$.
Model the rapid contraction as a switch to approximately constant systolic elastance $E_S$, with $C_s=1/E_S$ the systolic ventricular compliance. A sufficiently small aortic-valve resistance makes $p_{LV}\simeq p_a$ after a short <pressure>-equalization transient. The systolic pressure must be high enough to open the valve, $E_S(V^+-V_0)>p_a^+$. During that transient neglect systemic runoff. Conservation of combined arterial and stressed ventricular volume gives
$$
C_ap_a^++V^+-V_0=(C_a+C_s)\widehat p_a,
$$
so
$$
\boxed{\widehat p_a\simeq\frac{C_ap_a^++V^+-V_0}{C_a+C_s}.}
$$
This is an approximation across a rapid transient, not a discontinuity at finite valve resistance. The subsequent common <pressure> obeys $R_c(C_a+C_s)\dot p_a\simeq-p_a$, giving
$$
\boxed{p_a(t)\simeq\widehat p_a\exp\left[-\frac{t}{R_c(C_a+C_s)}\right].}
$$
The source's $R_C$ in this expression denotes the same systemic resistance $R_c$.
At the end of firing, the remaining ventricular volume is $V_{\mathrm{end}}\simeq V_0+C_sp_a(\Delta t_F)$. The <stroke volume> is $V^+-V_{\mathrm{end}}$, hence
$$
\boxed{\mathrm{SV}\simeq V^+-V_0-\frac{C_s}{C_a+C_s}(C_ap_a^++V^+-V_0)\exp\left[-\frac{\Delta t_F}{R_c(C_a+C_s)}\right].}
$$
The assumptions are a closed <mitral valve> throughout ejection, negligible venous <pressure>, rapid valve equalization relative to systemic runoff, and nearly constant systolic elastance during the interval measured by $\Delta t_F$. These give the <ventricular ejection pressure and stroke-volume approximation>.
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