How to solve this d'Alembert's equation? the ODE:
$$y'y=1-x(y')^3$$
Can be written in the form of the d'Alembert's equation:
$$y=xf(y')+g(y')$$
$$=\dfrac{1}{y'}-x(y')^2$$
But after that, i don´t know how to get to the solution for $y$:
Can some one explain me the process?
 A: A d'Alembert differential equation (curiously not listed on wikipedia's list),
has linear solutions $y=f(p)x+g(p)$ whenever $p=f(p)$. Outside these solutions one can use $p=y'$ as local parameter in $x=X(p)$, then $Y(p)=y(X(p))$ which implies by the chain rule $$Y'(p)=y'(X(p))X'(p)=p·X'(p).$$ 
Differentiating the original equation for $p$ gives
$$
Y'(p)=f'(p)X(p)+f(p)X'(p)+g'(p).
$$
in this concrete case
$$
Y(p)=\frac1p-X(p)p^2\implies Y'=-\frac1{p^2}-2pX-p^2X'
$$
Now eliminate $X$ or $Y$ and their respective derivatives from this equation using the two previous identities,
$$
(p-f(p))X'(p)=f'(p)X(p)+g'(p)
$$
or
$$
(p+p^2)X'(p)=-2pX(p)-\frac1{p^2}
$$
which can be solved as a linear ODE.

For the homogeneous solution one gets
$$
(\ln(X_h))'=-\frac{2}{p+1}\implies X_h=\frac{C}{(p+1)^2}
$$
Varying the constant resp. applying the integrating factor gives
$$
((p+1)^2X)'=(p+1)^2X'+2(p+1)X
=-\frac{p+1}{p^3}
$$
so that finally
$$
X(p)=\frac{1}{(p+1)^2}\left(C+\frac{1}{p}+\frac{1}{2p^2}\right)
$$
Using the trinitial equation completes the parametrization,
$$
Y(p)=\frac1p-p^2X(p)=\frac1p-\frac{1}{(p+1)^2}\left(Cp^2+p+\frac{1}{2}\right)
$$
