How to find the solution to the ode? $$\frac{d}{dt}\frac{p'^2(t)}{4p(t)[1-p(t)]}=0$$
or equivantly,
$$\frac{d}{dt} [(\sqrt{p(t)})'^2+(\sqrt{1-p(t)})'^2]=0$$
Or
$$p'^2(t)-2p(t)p'^2(t)-2p(t)p''(t)+2p^2(t)p''(t)=0$$ It sounds quit hard. Even not a linear ode.
 A: Here's another approach (of course, made once I knew the final answer). Substitute $p(t) = q(t)^2$, so that $p'(t) = 2q(t)q'(t)$. Then the first integral becomes
$$
\frac{d}{dt} \frac{p'(t)^2}{4p(t)(1-p(t))} = \frac{d}{dt} \frac{q'(t)^2}{1-q(t)^2}.
$$
Integrating this yields
$$
q'(t)^2 + C q(t)^2 = C
$$
for some constant $C$. Now, $C$ could be positive, negative, or zero, leading to different solutions. If $C$ is zero, then of course $q'(t) = 0$ and so $q$ is a constant. If it's positive, it's interesting to let $C = \omega^2$ and $q(t) = f(\omega t)$, so that
$$
f'(t)^2 + f(t)^2 = 1.
$$
Finally, if $C$ is negative, let $C = - \rho^2$ and $q(t) = f(\rho t)$, so that
$$
f'(t)^2 - f(t)^2 = -1.
$$
A: One immediate integration is possible:
$$\frac{(p')^2}{4p[1-p]}=C_1.$$
From this equation, it is evident that $C_1$ and $p(1-p)$ must have the same sign. Let us assume that $C_1>0,$ and that $p\in(0,1).$ Then we can take the square root of both sides (positive or negative):
\begin{align*}
\frac{p'}{\sqrt{p(1-p)}}&=\pm 2\sqrt{C_1}\\
\frac{dp}{\sqrt{p(1-p)}}&=\pm 2\sqrt{C_1}\,dt\\
-2\arcsin\left(\sqrt{1-p}\right)&=\pm 2\sqrt{C_1}\,t+C_2.
\end{align*}
You can easily solve this for $p(t).$
On the other hand, if you assume that $C_1<0,$ then correspondingly you must have $p\in(-\infty,0)\cup(1,\infty).$ You can apply negative signs to both sides and then proceed as before.
