I have been trying to solve a differential equation but I'm shard stucked.
The equation is: $y = x(y')^2 - \frac{1}{y'}$
As it has Lagrange form, I do $y' = p$.
$y = xp^2 - \frac{1}{p}$
Taking differentials
$dy = p^2dx + [2xp + \frac{1}{p^2}]dp$
As $y'=p \rightarrow dy=pdx$
Resuming:
$pdx = p^2dx + [2xp + \frac{1}{p^2}]dp$
$pdx - p^2dx = [2xp + \frac{1}{p^2}]dp$
$[p - p^2]dx = [2xp + \frac{1}{p^2}]dp$
Okay, so... any idea of how to resolve this equation? I dont know how to follow...
$$(p - p^2)dx = (2xp + \frac{1}{p^2})dp\quad\text{is OK.}$$ $$\frac{dx}{dp}=-\frac{2x}{p-1}-\frac{1}{p^3(p-1)}$$ This is a first order linear PDE for the function $x(p)$. Solving leads to : $$x=\frac{1}{p(p-1)^2}-\frac{1}{2p^2(p-1)^2}+\frac{c}{(p-1)^2}$$ and with $\quad y=xp^2-\frac{1}{p} =\frac{p}{(p-1)^2}-\frac{1}{2(p-1)^2}+\frac{c\:p^2}{(p-1)^2}-\frac{1}{p}\quad$ the solution is on parametric form :
$$\begin{cases} x=\frac{1}{p(p-1)^2}-\frac{1}{2p^2(p-1)^2}+\frac{c}{(p-1)^2}\\ y=-\frac{1}{p}+\frac{p}{(p-1)^2}-\frac{1}{2(p-1)^2}+\frac{c\:p^2}{(p-1)^2} \end{cases}$$