Reduction of order to find the solution of the non-linear equation

Question

How can one use reduction of order to find the solution of the non-linear equation:

$$(1 + y^2) y'' + (y ')^3-y' = 0.$$

Can anyone provide some reference/link where I can get the theory related to these types of the non-linear equation?

Answer 1

You can bring the equation into the form $f(y')y''+g(y)y'=0$ which is once integrable. $$ \frac{y''}{1-y'^2}-\frac{y'}{1+y^2}=0 \\~\\ {\rm Artanh}(y')-\arctan(y)=C \\~\\ y'=\tanh(C+\arctan(y)) $$ While this is again a separable equation, the integral that results does not look like it has a symbolic solution.

Answer 2

$$(1 + y^2) y'' + (y ')^3-y' = 0.$$ Substitute $p=\dfrac {dy}{dx}$: $$(1 + y^2) p\dfrac {dp}{dy} =p(1-p^2)$$ $$p((1 + y^2) \dfrac {dp}{dy} -(1-p^2))=0$$ You have $p=0 \implies y=c$

For $1-p^2=0 \implies y=\pm x+c$

And the separable DE for $1-p^2 \ne 0$: $$\int \dfrac {dp}{1-p^2} =\int \dfrac {dy}{1+y^2}$$ You can easily integrate but after that as Lutz pointed out in his answer it's not going to be easy to integrate again the DE.