We are asked to reduce $y^"+y-y^{-3}=0$ using $X= \sin2x\frac{\partial}{\partial x}+y\cos2x\frac{\partial}{\partial y}$
I know we have to find the first prolongation of X and solve $X^1$F=0 using $X^1=\xi\frac{\partial}{\partial x}+\eta\frac{\partial}{\partial y}+\eta^1\frac{\partial}{\partial y^1}$ and the formula for $\eta^1$ I get $\eta^1$ = $-2y\sin2x - y'\cos{2x}$ Is this correct? I then attempted to find the independent invariants by solving $X^1$F=0 and solving the dx,dy equation I got $\beta_1 = \frac{y^2}{\sin2x}$ or $\frac{y}{\sqrt{\sin2x}}$ depending on how I integrated (where I put the 1/2) I solved the first form for y and substituted in the dx,dy' equation which I attempted to solve for $\beta_2$ but I ended up with something like $dy'=\frac{-2\beta_1\sin2x - \beta_1\cos^22x}{\sqrt{\beta_1\sin2x}}dx$. I don't know if I did this correctly but now I'm stuck. I know I need the second invariant to calculate $\frac{d\beta_2}{d\beta_1}$ which reduces the differential equation. Any help would be appreciated.
$y"(x)+y(x)-y^{-3}=0$ is an autonomous ODE of second order.
Let $y'(x)=f(y)$ then $y''(x)=f'(y)y'(x)=f'(y)f(y)$
$f'(y)f(y)+y-y^{-3}=0$ is a first order ODE
$f^2=-y^2-y^{-2}+C$
$y'(x)=f(y)=\sqrt{-y^2-y^{-2}+C}$
$x(y)=\int{(-y^2-y^{-2}+C)^{-1/2}}dy$