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.

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    $\begingroup$ How did this o.d.e. arise? Prima facie it looks difficult. $\endgroup$ Dec 19, 2022 at 19:45
  • $\begingroup$ Unfortunately, not from a textbook, so an analytical solution may not even be possible. It's a small part of a research problem I'm working on (PhD) $\endgroup$ Dec 19, 2022 at 20:16
  • $\begingroup$ Alas, Maple doesn't find a nice solution. i.imgur.com/eULobTp.png $\endgroup$ Dec 19, 2022 at 20:31
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    $\begingroup$ @Giraffes4thewin is an analytical solution necessary for your purposes? Depending on the research problem, you might be able to get away with merely establishing properties that solutions to the ODE must have. $\endgroup$ Dec 20, 2022 at 6:18
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    $\begingroup$ Look into the Abel equation of the second kind which this can be transformed into, a la @user3782 $ww'-w=y^2$. There are known solutions for similar equations, like $ww'-w=Ay^2-9/(625A)$ but I cannot find one for this particular eqn. $\endgroup$ Dec 20, 2022 at 19:27

2 Answers 2


$$y'(x)=\frac{1}{x'(y)}\\ \frac{dy'}{dy}=-\frac{x''}{(x')^2}\\ y''=\frac{dy'}{dy}\frac{dy}{dx}=-\frac{x''}{(x')^3}$$ 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 $y(x)$.

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    $\begingroup$ Thank you for your effort in answering this question. $\endgroup$ Dec 22, 2022 at 21:58
  • $\begingroup$ @Giraffes4thewin You are welcome! $\endgroup$ Dec 23, 2022 at 13:34

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

  • $\begingroup$ Could you go ahead please ? $\endgroup$ Jan 8 at 10:34
  • $\begingroup$ Thought it would simplify, but now find it does not. Apologies for the unintended mislead. The recasting as above is also so far not lucky. $\endgroup$
    – Narasimham
    Jan 8 at 22:16

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