Textbook Problem does not seem to chose right ODE to model tank with water

Question

The problem, taken from the Spanish version of Numerical Methods for Engineers by Chapra, states the following: A vertical cylindrical tank drops it's water level at the following rate

$$\frac{dy}{dt} = -k\sqrt{y}$$

where $t$ is time and $k$ is a constant that depends on the draining source; when $y(0) = 3$. Solving this ODE I get the following function (taking the positive root of $y$)

$$y = \bigg( \frac{2\sqrt{3} - kt}{2} \bigg)^2$$

This being a parabola, signifies that after a time $2\sqrt{3}/k$ the water level starts rising again. This last part is what I don't understand, since the problem never mentions water poured into the tank.

So basically the question is why the water level is rising again and what it means physically. Is the ODE wrong? Am i solving this the wrong way? Any light shed on this is highly appreciated

Answer 1

The subtlety here is that the solution $y(t) = \left( \frac{2\sqrt{3} - kt}{2} \right)^2$ is valid only for $t\le2\sqrt3/k$.

At $t=2\sqrt3/k$, the value of $y(t)$ as defined above is zero, and the differential equation asserts that $dy/dt$ will be zero at that moment -- which means $y$ must remain zero for all subsequent $t$. So the proper solution to the differential equation is: $$ y(t)=\begin{cases} \left( \frac{2\sqrt{3} - kt}{2} \right)^2&\text{if $t\le2\sqrt3/k$}\\ 0&\text{if $t>2\sqrt3/k$} \end{cases} $$

Answer 2

The solution to the ode is $kt + c = -2y^{1/2}$, which is only possible for real numbers when $kt + c \leq 0$. When you square both sides, you create extra solutions that weren't there.