# Water tank problem (ODE)

I'm getting stuck in a water tank problem modeled by an ODE. However, the latter seems to be rather difficult to understand. I guess it is not simply a subtraction of rate flow in minus rate flow out, in order to obtain the actual rate of change of the height.

A rectangular water tank has a horizontal square base of side $1\,\rm m$. Water is being pumped into the tank at a constant rate of $400\,\rm cm^{3}\, s^{-1}$. Water is also flowing out of the tank from an outlet in the base. The rate at which water flows out at any time $t$ seconds is proportional to the square root of the depth, $h\,\rm cm$, of water in the tank at that time.

Explain how the information given above leads to the differential equation.

$$\frac{dh}{dt}=0.04-0.01 \sqrt{h}$$

$400\,\frac{\rm cm^3}{\rm s}$ = $0.0004\,\frac{\rm m^3}{\rm s}$ and, since the base has area $1\,\frac{\rm m^2}{\rm s}$, the water pumped in at any given moment increases the height by $.04\,\rm cm$.