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 present this way for solution your answer. Following the below way 
                             dh/dt=-k√(h,)
You can solve this equation very easy. Divide both side of equation to √(h,). In this case, equation is changed to a separated of variable. By integrating both of them relative to separation variable and using initial condition, at which h(t=0)=h_0 , you can find integration constant, c=(h_0 )^1/2 and finally its answer:
h=(-1/2*kt+h_0^(1/2))^1/2
This equation shows the h versus t variation. By the way, you can draw it to linearity with h^2 versus t.