Can anyone solve this hard differential equation involving a derivative squared? I have been trying to solve this diff. equation for quite some time now but haven't been able to do it correctly. It describes the lost height of the water $h$ at a certain time $t$ of a leaking reservoir. I have obtained this equation by using Torricelli's law as well as the law of continuity. At $t=0$, the seal is removed and the reservoir starts leaking. Moreover, $A_O > A_G$ and to be less precise it can be assumed $A_O >> A_G$, however, I would like to solve it as precisely as possible. The equation is:
$$\frac{2gA_O^2}{A_O^2 - A_ G^2}h(t)+\frac{A_O^2}{A_G^2}\bigg(\frac{dh(t)}{dt}\bigg)^2 - \frac{2gA_O^2}{A_O^2 - A_G^2}H_0=0$$   with $h(0)=0$. Additionally, $h(t)$ is bounded above by $H_0$, is a strictly increasing function and thus also $\lim_{t \to \inf} h(t)=H_0$
The constants:


*

*$A_O$ is the surface of the top of the reservoir.

*$A_G$ is the surface area of the hole.

*$g$ is the gravitational acceleration.

*$H_0$ is the height of the water at t=0.
The simplified formula would be:
$$\lambda h(t)+\mu\left(\frac{dh(t)}{dt}\right)^2-\lambda H_0=0$$
And just to be clear $\bigg(\frac{dh(t)}{dt}\bigg)^2$ is simply the first derivative squared not the second derivative.
I really hope someone can help me solve this!
If there is anything unclear or you want more information please ask I will check this post regularly. Thanks in advance!
 A: First, write $h(t)=H_0+f(t)$ to get a slightly simpler equation
$$
\frac{2gA_O^2}{A_O^2 - A_ G^2}f(t)+\frac{A_O^2}{A_G^2}\bigg(\frac{df(t)}{dt}\bigg)^2=0.
$$
Writing $r=A_O/A_G$, we can simplify the constants and find
$$
\frac{2g}{r^2 - 1}f(t)+\bigg(\frac{df(t)}{dt}\bigg)^2=0.
$$
If $f'(t)=0$, the equation tells you $f(t)=0$.
I'll assume $r>1$ so that the equation forces $f\leq0$.
This holds as $A_0>A_G$.
The other case will be similar but I will stick to this choice as in the updated question.
If $f'(t)\neq0$, you get
$$
f'(t)
=
\pm\sqrt{-\frac{2g}{r^2 - 1}f(t)}
$$
or
$$
\frac{df}{-\sqrt{\frac{2g}{r^2 - 1}f}}
=
\pm dt.
$$
Integrating gives
$$
\sqrt{\frac{r^2 - 1}{8g}}
\sqrt{-f}
=
t_0\pm t
$$
for some $t_0$, so
$$
f(t)
=
-\frac{8g}{r^2 - 1}
(t_0\pm t)^2.
$$
As the time difference is squared anyway, you can rewrite this general solution as
$$
f(t)
=
-\frac{8g}{r^2 - 1}
(t-t_0)^2.
$$
This gives you two kinds of solutions, and for $r<1$ you will get something similar.
Be careful with solutions when $f=0$ (which is at $t=t_0$).
There the solution fails to be unique because the function can stop at the zero level.
For example,
$$
f(t)
=
\begin{cases}
-\frac{8g}{r^2 - 1}
t^2,&t<0\\
0,&0\leq t\leq 1\\
-\frac{8g}{r^2 - 1}
(t-1)^2,&t>1
\end{cases}
$$
is a solution.
Your definition of $H_0$ requires that $f(0)=0$, as $f=0\iff h=H_0$.
With your update I think you mean $h(0)=0$ instead of $h(0)=H_0$.
Let us then see what happens with your added assumption that $h(t)\leq H_0$, $h$ is strictly increasing and $\lim_{t\to\infty}h(t)=H_0$.
With my notation this gives $f(t)\leq0$, $f'(t)\geq0$, and $\lim_{t\to\infty}f(t)=0$.
The updated initial condition seems to be $f(0)=-H_0$.
Let's start with
$$
f(t)
=
-\frac{8g}{r^2 - 1}
(t-t_0)^2.
$$
Putting in the initial condition gives
$$
-H_0
=
f(0)
=
-\frac{8g}{r^2 - 1}
t_0^2,
$$
so
$$
f(t)
=
-a
(t-\sqrt{H_0/a})^2,
$$
where $a=\frac{8g}{r^2 - 1}$.
As $f$ should be increasing, we have
$$
f(t)
=
\begin{cases}
-a
(t-\sqrt{H_0/a})^2
,&
0\leq t\leq \sqrt{H_0/a}\\
0
,&
t>\sqrt{H_0/a}.
\end{cases}
$$
The function is increasing but not strictly increasing; it reaches it's final value in finite time.
Without the increasing assumption the solution can decide to "bounce back down" starting at any time after $t_0=\sqrt{H_0/a}$.
A: You could put it in the form 
$$\left(y’\right)^2 = ky $$
where $y =h-H_0$ and $k=-\lambda/\mu$.
With assumptions about various signs, you might even go for
$$y’=cy^{1/2}$$
