# Differential Equation Describing the Flowing of Water Out of a Tank

I have the differential equation $${{dh} \over {dt}} = - K\sqrt h$$

describing water flowing out of the bottom of a tank of uniform cross-sectional area under the action of gravity. $h(t)$ is the water depth at time t with ${h_0}$ being the initial depth. K is a positive constant.

I have no idea how to solve this. I only know how to separate variables to solve DEs. How would I solve this one? This is one of the starred questions in the book. I can't find another example like this. Thanks!

You can still solve this with separation of variables. divide both sides by $\sqrt{h}$ and you have a separated equation.
• I did that but I got that $t = - {{2\sqrt h } \over K}$ so when I plug in h = 0, that means the tank empties instantaneously... What am I doing wrong? – SumMathGuy Feb 24 '16 at 18:23
• After dividing by $\sqrt{h}$ and integrating both sides we have the equation $2\sqrt{h} = -Kt + C$ where C is a constant. solving for h here gives $h = ((-Kt+C)/2)^2$ – CrazyIvan Feb 24 '16 at 18:27
• @SumMathGuy You have to integrate using proper limits. You will get $t=-\frac{2\sqrt h}{K} + c$, where $c$ has to be found using the initial condition given. – GoodDeeds Feb 24 '16 at 18:27
• I did get that but I don't have an initial condition. As I said above, it just says ${h_0}$ so I can't solve for c. Ivan, I am solving for t not for h. – SumMathGuy Feb 24 '16 at 18:33
• h(0) = $h_0$ so when we plug in 0 for t we can solve for c. in particular $c = 2\sqrt{h_0}/K$. – CrazyIvan Feb 24 '16 at 18:35