TITLE: Non constant periodic solutions to $y''=y'-2(y')^2-3y+4y^2$
QUESTION [6 upvotes]: Does the equation $y''=y'-2(y')^2-3y+4y^2$ have non constant periodic solutions?

So far I transformed this equation into a first order ODE system by setting $x=y'$. This is what we saw in class:

The ODE $y'=v(y)$ has no non constant periodic orbits if there exists a function $\mu$ such that $\omega = \mu \sum v_i dy_i$ is exact.

I'm struggling to find $\mu$ and I'm wondering if there is a better method than just trial and error. I've tried rearranging the equation too and maybe isolate either x or y, but it didn't get me anywhere.

REPLY [2 votes]: I think I finally got it.
So first we divide everything by $y$. And then do a change of variables $u=y'/y$. So we have:
$u'+u^2=u'-2y'u-3+4y \iff u^2=-2y'u-3+4y \iff 2y'u=4y-u^2-3$
Define $f:=4y-u^2-3$ and $g:= 2u$. The form $\omega = fdy -gdu$ is not exact. But we can find an integrating factor.
Let $h:= \frac {\partial f} {\partial y} + \frac {\partial g} {\partial u} =6$. Since $\frac {\partial } {\partial y} (h/g) = 0$, we can find an integrating factor $\mu$ such that $\mu_u'=-(h/g) \mu$ $\implies \mu = \frac 1 {u^3}$. The form $\mu \omega$ is then exact.
And so, we can find the first integral $F=-\frac 2 {u^2} y+\frac 3 {2u^2}- \ln(u)$ (double check please, I'm bad at calculations)
Since we can find the first integral, the equation has no non constant periodic solutions.