Solving $2x(t)(x'(t))^3-(x'(t))^2=C$ I have differential equation:
$$2x(t)(x'(t))^3-(x'(t))^2=C$$
where $C$ is some constant and $x(0)=x(1)=0$.
I've solved this equation for $C=0$, then $x(t)=0$, but what about other $C$?
 A: This is quite a nice equation. It allows a solution, if we consider $t(x)$ instead of $x(t)$:
$$2x \frac{1}{t'^3}-\frac{1}{t'^2}=C$$

$$C t'^3+t'-2x=0 \tag{1}$$

Introducing new function:
$$f(x)=t'(x)$$

$$C f^3+f-2x=0 \tag{2}$$

But this is just a cubic equation! It has an explicit solution in radicals or in trigonomentric functions. Thus, the problem is solved.
In particular, if $C=0$ then:
$$f(x)=2x$$
$$dt=2xdx$$
$$x = \pm \sqrt{t+C_1}$$
As was pointed out by tired in the comments.

Now let's study this further. The discriminant of (2) is:
$$\Delta = -4 C(1+ 27 C x^2)$$
If $C \neq 0$ then $\Delta \neq 0$ for general $x$. However, the discriminant can change sign for particular values of $x$. So we might get one real and two complex roots ($\Delta < 0$) or three real roots ($\Delta > 0$) depending on $x$.

However, there's a very convenient method of expressing the roots of a cubic equation in hypergeometric form. I unfortunately do not remember the reference where the method is introduced, it's a rather recent paper, I will try to find it later.
For equation:
$$y^3+3p y+2q=0$$
The three roots are:
$$y_1 = \frac{2q}{p} {_2 F_1} \left(\frac{1}{2},\frac{2}{3};\frac{1}{2};1+ \frac{q^2}{p^3}  \right)$$
$$y_{2,3} = -\frac{2q}{3p} {_2 F_1} \left(\frac{2}{3},\frac{4}{3};\frac{3}{2};\frac{1}{2} \left(1 \pm \sqrt{1+ \frac{q^2}{p^3}} \right)  \right)$$
In our case we have:

$$f_1(x)=-6 x~ {_2 F_1} \left(\frac{1}{2},\frac{2}{3};\frac{1}{2};1+ 27 C x^2  \right)$$
$$f_{2,3}(x) = 2x ~{_2 F_1} \left(\frac{2}{3},\frac{4}{3};\frac{3}{2};\frac{1}{2} \left(1 \pm \sqrt{1+ 27 C x^2} \right)  \right)$$

Depending on $C$ and $x$ either one of these roots can be real or complex (their form doesn't tell us which one).
What's the advantage of Hypergeometric functions? They can be integrated explicitly, usually resulting in Hypergeometric functions as well. So we can in principle obtain $t(x)$ in more or less closed form.
