A problem about fluid flow A tank with vertical sides has a square cross-section of area $4$ square feet. Water is leaving the tank through an orifice of area $5/3$ square inches. If the water level is initially $2$ feet above the orifice, how to find the time required for the level to drop $1$ foot? 
If water also flows into the tank at the rate of $100$ cubic inches per second, then how can we show that the water level above the orifice approaches the value $(25/24)^2$ feet above the orifice, regardless of the initial water level? 
 A: I'm doing the exercise too.
Questions 1 is simple. Denote the area of orifice $A_0$ and we'll have $ \frac{dV}{dt}=- c\sqrt{2gy}A_0$ and $\frac{dV}{dy}=A(y)=4$.use the chain rule we have $$\frac{dy}{dt}=\frac{dy}{dV} \frac{dV}{dt} = -\frac{c \sqrt{2gy}A_0}{A(y)}$$ cause $c\sqrt{2g}A_0 /A(y)$ is constant ,denote it by $K$ and rewrite the formula we get $$\frac{y'}{\sqrt{y}}=-K$$Integrate it with the data given $$\int_2^1 \frac{y'}{\sqrt{y}}=-K \int_0^xdt$$
.The only thing have to take notice is the unit of $A_0$ is "square inches" but the one of A(y) is "square feet".
Question 2. I get a idea to solve it.
Denote the volume of water flowed in by $Vi$ ,we have  $\frac{dV}{dt}=V_i-c\sqrt{2gy}A_0$,$\frac{dV}{dy}=A_y$ ,
use the chain rule to get $$\frac{dy}{dt}=\frac{dy}{dV} \frac{dV}{dt} = \frac{V_i - c \sqrt{2gy}A_0}{A_y}$$ let's say $y'=n-m\sqrt{y}$ ($n$ and $m$ are both positive)for simplicity. And  let $r= 1/2$ and $$g=y^{1-r}=\sqrt{y},g'=(1-r)y^{-r}y'$$ we have
  $$g'=(1-r)(ng^{-1}-m)=\frac{1}{2}\frac{n-mg}{g}$$ it's seperatable.Rewrite it and get$$(\frac{1}{m}+\frac{n}{m^2} \frac{m}{mg-n})g' = - \frac{1}{2}$$ Solve this equation we have an implicity formula:$$\frac{g}{m} + \frac{n}{m^2}\log|mg-n|=-\frac{t}{2}+C  (or)  \frac{\sqrt{y}}{m} + \frac{n}{m^2}\log|m\sqrt{y}-n|=-\frac{t}{2}+C$$ for some constant $C$.
The question is $y \to (\frac{25}{24})^2$ as $t \to + \infty$,notice  the RHS approach to $-\infty$,so the LHS  should approach to $-\infty$ too. Because the first term of the LHS is always opsitive.The only way to achieve this is $\log|m\sqrt{y}-n| \to - \infty$ means $m\sqrt{y}-n \to 0 $ as $t \to + \infty$ ,so $y$ approaches to $(\frac{n}{m})^2=(\frac{25}{24})^2$  as $t \to + \infty$. Check the answer we'll see $$K + -\infty = -\infty + C$$
