What is the equation for when a project will end given this formula for how fast work is done? I want to calculate the rate at which work gets done on a project. On each week some work is done and some new work is generated. The new work generated is directly proportional to the amount of work remaining in the project. **Work done on project** $dW = (cW - R) dt$ W = work remaining R = rate of work c = constant representing churn rate How do I calculate when the project ends? **EDIT** _Edited to remove original sign error where I had: $dW = (R - cW) dt$_

The general solution of this ordinary linear differential equation is

$$W=\frac Rc+a\mathrm e^{ct}\;.$$

Thus, if the initial workload is above $R/c$, work will pile up exponentially ($a\gt0$), whereas if the initial workload is below $R/c$, it will decrease exponentially ($a\lt0$), but the rate won't exceed $R$ before the work becomes $0$. The time $T$ to finish as a function of the initial workload is given by

$$W_0=W(t=0)=\frac Rc+a\;,$$

$$a=W_0-\frac Rc\;,$$

$$0=W(t=T)=\frac Rc+a\mathrm e^{cT}=\frac Rc+\left(W_0-\frac Rc\right)\mathrm e^{cT}\;,$$

$$\left(\frac Rc-W_0\right)\mathrm e^{cT}=\frac Rc\;,$$

$$\mathrm e^{cT}=\frac{\frac Rc}{\frac Rc-W_0}=\frac{1}{1-\frac{cW_0}R}\;,$$

$$cT=\log\frac{1}{1-\frac{cW_0}R}=-\log\left(1-\frac{cW_0}R\right)\;,$$

$$T=-\frac{\log\left(1-\frac{cW_0}R\right)}c\;,$$

where $\log$ denotes the natural logarithm.