Nonlinear second-order ODE $yy'' - (y')^{2} = y^4$ I have the following ODE to solve.
$$
yy'' - (y')^{2} = y^4
$$
I tried to substitute $y'$ by $v$, and then I get the following:
$$
yv' - v^{2} = y^4.
$$
I can't go further. I can't see what I'm supposed to do in order to solve it.
I saw a solution involving Bessel function. But, is it possible to transform the first ODE into an exact, linear, or Bernoulli equation?
Any hint, please. Thanks.
 A: The trick is using $v = y' = \frac{dy}{dt}$ but expressing the equation a different way
$$y'' = \frac{dv}{dt} = \frac{dy}{dt}\frac{dv}{dy} = v\frac{dv}{dy} $$
Thus
$$ yv\frac{dv}{dy} = v^2 + y^4 $$
From here, let $zy = v^2$ so $$z + y\frac{dz}{dy} = 2v\frac{dv}{dy} = \frac{2}{y} \left( zy + y^4 \right) = 2z + 2y^3$$
$$\frac{dz}{dy} = \frac{z}{y} + 2y^2$$
This a linear first-order ODE that you can solve
A: If we set $y(t)=\exp\left(f(t)\right)$ then we have $f(t)=\log y(t)$ and:
$$ f''(t) = \frac{y y''-(y')^2}{y^2}\tag{1}$$
hence the original ODE is mapped to:
$$ f''(t) = \exp(2 f(t)) \tag{2}$$
and by multiplying both sides by $2f'$ then integrating we get:
$$ \left(f'(t)\right)^2 = e^{2\, f(t)} + C\tag{3} $$
that is a separable ODE. To solve it, we just need to find a primitive for $\frac{x}{\sqrt{e^{2x}+C}}$.
Assuming $C=0$ that task is pretty easy to accomplish. If $C\neq 0$, a dilogarithm is involved.
A: Divide by $(1+ y^{'2})= \sec^2 \phi, $ 
and let $ \cos^2 \phi = U ;$
It reduces to the form:
$$ \dfrac{d(log\; U)}   {d\,( log\, y^2 )  } = (y^4 U^2 - U+1) $$
We can proceed with it further.
