Show that the depth of water in the tank at any time satisfies the below differential equation

Question

I saw a similar problem from Standford, but I couldn't figure out how to apply Standford's answer to this problem.

Answer 1

The volume $V(t)$ of water in the tank at time $t$ is given by

$$V(t)=\frac{\pi h^3(t)}{3}\,\tan^2(\phi)$$

Hence, the rate of change in $V(t)$, $\frac{dv(t)}{dt}=\overbrace{k}^{\text{Inward flow rate}}-\overbrace{\alpha \sqrt{h(t)}}^{\text{Outward flow rate}}$, is given by

$$\begin{align} \frac{dV(t)}{dt}&=\pi h^2(t)\tan^2(\phi)\frac{dh(t)}{dt}\\\\ &=k-\alpha \sqrt {h(t)} \end{align}$$

whereupon solving for $\frac{dh(t)}{dt}$ yields

$$\bbox[5px,border:2px solid #C0A000]{\frac{dh(t)}{dt}=\frac{k-\alpha \sqrt{h(t)}}{\pi h^2(t)\tan^2(\phi)}}$$

as was to be shown!