Problem source

\begin{questionparts}
\item At time $t=0$ a tank contains one unit of water. Water flows out of the tank at a rate proportional to the amount of water in the tank. The amount of water in the tank at time $t$ is $y$. Show that there is a constant $b < 1$ such that $y=b^{t}.$ 
\item Suppose instead that the tank contains one unit of water at time $t=0,$ but that in addition to water flowing out as described, water is added at a steady rate $a>0.$ Show that 
\[
\frac{\mathrm{d}y}{\mathrm{d}t}-y\ln b=a,
\]
and hence find $y$ in terms of $a,b$ and $t$. 
\end{questionparts}

Solution source

\begin{questionparts}
\item Since water flows out a rate proportional to the water in the tank we must have $\dot{y} = -ky$, ie $y = Ae^{-k t}$. Since $t = 0, y = 1$ we have $y = e^{-kt} = (e^{-k})^t$, so call $b = e^{-k}$ and we have the result. (Since $k > 0 \Rightarrow b < 1$

\item Notice that 
\begin{align*}
&& \dot{y} &= -\underbrace{ky}_{\text{flow out}} + \underbrace{a}_{\text{flow in}} \\
&&&= y\ln b + a \\
\Rightarrow && \dot{y} - y \ln b &= a \\
\\
\text{CF}: && y &= Ae^{\ln b t}  = Ab^t\\
\text{PI}: && y &= -\frac{a}{\ln b}  \\
t = 0, y = 1: && 1 &= A-\frac{a}{\ln b} \\
\Rightarrow && y &= \frac{a}{\ln b} \left ( b^t - 1 \right)+b^t
\end{align*}
\end{questionparts}