Solve the ODE $y'=(yy')'-y$

Question

Solve, or make any progress in reducing the ODE

\begin{equation} y' = (yy')'-y \end{equation}

on $x \in [0,1]$. I have tried reducing the order of this ODE by introducing the coordinate $y'(x)=p(y)$, so that $y''(x)=pp'(y)$ and I obtain the first order ODE

\begin{equation} p'(y) = \frac{p-p^2+y}{yp} \end{equation}

but can't seem to find a solution to this either. Thanks.

Answer 1

$$y'(x)=\frac{1}{x'(y)}\\ \frac{dy'}{dy}=-\frac{x''}{(x')^2}\\ y''=\frac{dy'}{dy}\frac{dy}{dx}=-\frac{x''}{(x')^3}$$ Now, substitute this into the DE: $$(x')^2=x'-yx''-y(x')^3\\ x'(y)=r(y)\\ r^2=r-yr'-yr^3\\ r'=\frac{r-r^2-yr^3}{y}\\ r(y)=z(y)y\\ z'=-z^2-y^2z^3\\ \frac{z'}{z^3}=-\frac{1}{z}-y^2\\ w(y)=\frac{1}{z(y)}\\ ww'=w+y^2$$ As pointed out by @Eli Bartlett, this is an Abel DE of the second kind and doesn't have an elementary solution and I don't think it is even analytically solvable.

Edit: The DE can also be reduced to an Abel DE of the first kind: $$27y^6e^yu^3+27y^4e^{y/3}u'+3y^2+18y+1=0$$ where $u(y)=\left(z(y)+\frac{1}{3y^2}\right)e^{-y/3}$ which is motivated by completing the cube on the RHS to get $\left(z(y)+\frac{1}{3y^2}\right)^3=\ldots-\frac{z'(y)}{y^2}$ and making the substitution $f(y)=z(y)+\frac{1}{3y^2}$ and then making another substitution $f(y)=g(y)h(y)$ and choosing $g(y)$ to eliminate the $f^1(y)$ term by solving a DE in $g(y)$ that gives $g(y)=e^{y/3}$.

This DE is of the form: $$u_y'=P(y)u^3(y)+Q(y)$$ Letting $s=\int P(y)dy$ puts the DE in canonical form. $$u_s'=u^3(s)+\frac{Q(s)}{P(s)}\\ u_s'=u^3(s)+\phi(s)$$ which can be solved analytically as per Wikipedia and I found this paper that states the solution method.

If practical, after solving this DE, we would get: $u=f(s)$ and substitute $s=\int P(y)dy$ to get $u=f\left(\int P(y)dy\right)$ and then substitute $u=\left(\frac{x'(y)}{y}+\frac{1}{3y^2}\right)e^{-y/3}$ to get $x'(y)=yf\left(\int P(y)dy\right)e^{y/3}-\frac{1}{3y}$ and integrate to get $x(y)$, then invert using series reversion to solve for $y(x)$.

Answer 2

HINT: $Let ~ v= y^2/2,~ v'= yy'= yp,~ \to \dfrac{dv}{dx}-\dfrac{v}{p(x)}=p(x)~; \tag 1 $

Another trial

$$ Let~ \frac{y^2}{2}=u, ~ yy'= u',~ (yy')'=u'' = yy''+y^{'2} = y+y' \text{(given)} $$ $$ y''+\frac{y'}{y}(y'-1) =1 ~\tag2 $$

which is still not amenable..

2nd order NL ODE