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First, I should clarify that for bijective $y(x)$ and $\frac{dx}{dy}\neq0$: $$\frac{dy}{dx}=\left(\frac{dx}{dy}\right)^{-1}\\ \frac{d^2y}{dydx}=-\left(\frac{dx}{dy}\right)^{-2}\frac{d^2x}{dy^2}\\ \frac{d^2y}{dx^2}=\frac{dy}{dx}\frac{d^2y}{dydx}=-\left(\frac{dx}{dy}\right)^{-2}\frac{d^2x}{dy^2}\frac{dy}{dx}=-\left(\frac{dx}{dy}\right)^{-3}\frac{d^2x}{dy^2} $$ I was recently trying to solve an abstract ODE question: $$\frac{d^2y}{dx^2}+a_0(y)\frac{dy}{dx}+a_1(y)=0 $$ aka second-order autonomous ODE.

I can transform it into an ODE in $x$: $$-\frac{d^2x}{dy^2}+a_0\left(\frac{dx}{dy}\right)^2+a_1\left(\frac{dx}{dy}\right)^3=0 $$ which can be transformed to a non-linear first-order ODE by substituting $x_0(y)=\frac{dx}{dy}$: $$\boxed{-\frac{dx_0}{dy}+a_0x_0^2+a_1x_0^3=0} $$ The question is: can these nonlinear ODEs be solved to solve the — I would expect — much harder second-order ODEs?

Substituting $x_0=e^{x_1(y)}$ and finding $x_1(y)$ such that $-\frac{dx_0}{dy}+\frac{d}{dy}(x_0^3x_1)$ is equal to the LHS don't work. My next idea was completing the cube so I can have $(x_0+x_1(y))^3$ to make a substitution and since the linear terms are missing, I will have two extra $x_0$-linear terms which will be duable via the substitution. However, I have an extra $x_2(y)$ which makes the ODE non-Bernoulli. $$\frac{-1}{a_1}\frac{dx_0}{dy}+\frac{a_0}{a_1}x_0^2+x_0^3=0\\ (x_0+x_1)^3=x_0^3+3x_1x_0^2+3x_1^2x_0+x_1^3\\ 3x_1=\frac{a_0}{a_1}\\ \frac{-1}{a_1}\frac{dx_0}{dy}+\left(x_0+\frac{a_0}{3a_1}\right)^3-\frac{a_0^2}{3a_1^2}x_0-\left(\frac{a_0}{3a_1}\right)^3=0\\ x_1=x_0+\frac{a_0}{3a_1}\\ \frac{dx_0}{dy}=\frac{dx_1}{dy}-\frac{d}{dy}\left(\frac{a_0}{3a_1}\right)\\ \frac{-1}{a_1}\frac{dx_1}{dy}+x_1^3-\frac{a_0^2}{3a_1^2}x_1=\frac{-1}{a_1}\frac{d}{dy}\left(\frac{a_0}{3a_1}\right)-\frac{2a_0^3}{27a_1^3}\\ \frac{dx_1}{dy}-a_1x_1^3+\frac{a_0^2}{3a_1}x_1=\frac{d}{dy}\left(\frac{a_0}{3a_1}\right)+\frac{2a_0^3}{27a_1^2} $$ All I know is that the equation is solvable if: $$a_0=0\vee a_1=0\vee\frac{d}{dy}\left(\frac{a_0}{3a_1}\right)+\frac{2a_0^3}{27a_1^2}=0\\ \frac{da_0}{dy}a_1-\frac{da_1}{dy}a_0+\frac23a_0^3=0\\ \frac{da_1}{dy}-\frac{1}{a_0}\frac{da_0}{dy}a_1=\frac{2}{3}a_0^2 $$ That is: $$\boxed{a_1=\frac23a_0\int a_0dy}$$ Otherwise, I can reduce equations of this form, i.e $$\frac{dx_1}{dy}+a_2x_1+a_3x_1^3=a_4$$ to ones with a non-zero constant instead of $a_4$: $$\frac{1}{a_4}\frac{dx_1}{dy}+\frac{a_2}{a_4}x_1+\frac{a_3}{a_4}x_1^3=1\\ x_2=\frac{x_1}{a_4}\\ \frac{dx_2}{dy}=\frac{1}{a_4}\frac{dx_1}{dy}-\frac{1}{a_4}\frac{da_4}{dy}x_2\\ \frac{dx_2}{dy}+\left(a_2+\frac{1}{a_4}\frac{da_4}{dy}\right)x_2+a_3a_4^2x_2^3=1 $$ That is: $$\boxed{\frac{dx_2}{dy}+a_5x_2+a_6x_2^3=1} $$ This is the closest I can get the equation to Bernoulli form. Now, the question is: what's the relation between the Bernoulli form (aka, the homogeneous case; when the RHS $=0$) and the equation of interest (when the RHS $=$ non-arbitrary non-zero constant)?

Edit: I discovered a way to reduce the problem to solving a simpler cubic ODE.

We start at: $$-\frac{dx_1}{dy}+a_0x_1^2+a_1x_1^3=0 $$ and we substitute $x_1=x_3(y)x_4(y)$: $$-x_3\frac{dx_4}{dy}-\frac{dx_3}{dy}x_4+a_0x_3^2x_4^2+a_1x_3^3x_4^3=0\\ -\frac{1}{a_1x_3^2}\frac{dx_4}{dy}-\frac{1}{a_1x_3^3}\frac{dx_3}{dy}x_4+\frac{a_0}{a_1x_3}x_4^2+x_4^3=0 $$ Now, we choose $x_3$ such that the cubic is perfect upto some $x_5(y)$, i.e $$\left(\frac{a_0}{3a_1x_3}\right)^2=-\frac{1}{3a_1x_3^3}\frac{dx_3}{dy}\\ \frac{1}{x_3}\frac{dx_3}{dy}=-\frac{a_0^2}{3a_1}\\ x_3=e^{-\int\frac{a_0^2}{3a_1}dy} $$ Thus, we have: $$-\frac{1}{a_1x_3^2}\frac{dx_4}{dy}+\left(x_4+\frac{a_0}{3a_1x_3}\right)^3=\left(\frac{a_0}{3a_1x_3}\right)^3\\ x_5=x_4+\frac{a_0}{3a_1x_3}\\ \frac{dx_4}{dy}=\frac{dx_5}{dy}-\frac{d}{dy}\left(\frac{a_0}{3a_1x_3}\right)\\ -\frac{1}{a_1x_3^2}\frac{dx_5}{dy}+x_5^3=\frac{1}{a_1x_3^2}\frac{d}{dy}\left(\frac{a_0}{3a_1x_3}\right)+\left(\frac{a_0}{3a_1x_3}\right)^3 $$ That is: $$\boxed{a_7\frac{dx_5}{dy}+x_5^3+a_8=0}$$

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  • $\begingroup$ Let $\kappa=1/f$. The equation can be written $(f'+a_1)f =a_3-a_2\xi$. Still nonlinear, but much neater. $\endgroup$ Commented Jul 3, 2022 at 17:07
  • $\begingroup$ @eyeballfrog This is exactly the form in JJacquelin's answer. However, this is a more straightforward substitution. $\endgroup$
    – user955932
    Commented Jul 3, 2022 at 17:13

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Below this is not an answer but a comment too long to be edited in the answers section.

If I well understand, you are trying to solve : $$\ddot\xi+a_1(\xi)\dot\xi+a_2(\xi)\xi=a_3(\xi)$$ In fact I am surprised by your approach to reduce the second order ODE to a first order ODE.

This is an ODE of autonomous kind. So why not trying the classical method of order reduction ?

$$\frac{d^2\xi}{d\eta^2}+a_1(\xi)\frac{d\xi}{d\eta}+a_2(\xi)\xi=a_3(\xi)$$

Note : On can simplify to $\quad \frac{d^2\xi}{d\eta^2}+a_1(\xi)\frac{d\xi}{d\eta}+a_4(\xi)\xi=0\quad$ with $\quad a_4(\xi)= a_2(\xi)\xi-a_3(\xi)$.

Change of function to reduce the order : $$\frac{d\xi}{d\eta}=u(\xi) \quad\implies\quad \frac{d^2\xi}{d\eta^2}=\frac{du}{d\xi}\frac{d\xi}{d\eta}=u\:\frac{du}{d\xi}$$

$$u\:\frac{du}{d\xi}+a_1(\xi)\:u=a_3(\xi)-a_2(\xi)\xi$$ $$u\:\frac{du}{d\xi}+a_1(\xi)\:u=a_4(\xi)$$ This is a nonlinear first order ODE to be solved for $u(\xi)$. Supposing that one can solve it the problem is far to be fully solved. One will have to integrate $\eta=\int \frac{d\xi}{u(\xi)}$ and finally invert $\eta(\xi)$ for $\xi(\eta)$.

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  • $\begingroup$ Thanks a lot, Jacquelin. The purpose of my approach was to get the equation as close as possible to a Bernoulli one. I did miss the classic order reduction technique (I am really not familiar with the terminology and classification of ODEs) but I do realise this equation is easier to solve than the cubic. One thing maths always taught me is that sometimes you gotta scrap the idea and start all over again, maybe there is an easier approach. $\endgroup$
    – user955932
    Commented Jul 3, 2022 at 13:24
  • $\begingroup$ You are welcome. I don't know if the method I propose is easier. I am afraid that it will not be easy in general. Often It will fail due to the final integration and invertion. $\endgroup$
    – JJacquelin
    Commented Jul 3, 2022 at 14:57

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