<s>[INST] Solving a nonlinear periodic ODE Is it possible to solve the below ODE for arbitrary real values of $c$?
$$2 (\frac{d \theta}{dx})^2 [\sin(2\theta) -c \cos(2 \theta)]-[\cos(2\theta) +c \sin(2 \theta)]\frac{d^2}{dx^2}\theta=0,$$
where $\theta=\theta(x)$. Also, $\theta$ itself does not have meaning, but ($\cos(2\theta),\sin(2\theta)$) are important and need to be continuous. There is no boundary condition since $x$ is not constrained by a boundary but goes to infinity.  It means that the solution must cover the range $(-\infty,\infty)$.
 [/INST] $$2 (\frac{d \theta}{dx})^2 [\sin(2\theta) -c \cos(2 \theta)]-[\cos(2\theta) +c \sin(2 \theta)]\frac{d^2}{dx^2}\theta=0$$
Let $y=\frac{d \theta}{dx}=\frac{d \theta}{dy}\frac{dy}{dx}\quad\implies\quad \frac{dy}{dx}=y\frac{dy}{d \theta}$
$$2 y^2 [\sin(2\theta) -c \cos(2 \theta)]-[\cos(2\theta) +c \sin(2 \theta)]\frac{dy}{dx}=0$$
$$2 y^2 [\sin(2\theta) -c \cos(2 \theta)]-[\cos(2\theta) +c \sin(2 \theta)]y\frac{dy}{d\theta}=0$$
$$\frac{dy}{y}=2\frac{\sin(2\theta) -c \cos(2 \theta)}{\cos(2\theta) +c \sin(2 \theta)}d\theta$$
$$\ln|y|=2\int \frac{\sin(2\theta) -c \cos(2 \theta)}{\cos(2\theta) +c \sin(2 \theta)}d\theta=-\ln|\cos(2\theta) +c \sin(2 \theta)|+\text{constant}$$
$$\frac{d \theta}{dx}=y=\frac{C_1}{\cos(2\theta) +c \sin(2 \theta)}$$
$$C_1x=\int \big(\cos(2\theta) +c \sin(2 \theta)\big)d\theta$$
$$C_1x+C_2=\frac12\big(\sin(2\theta) -c\: \cos(2 \theta)\big)$$
Then there is no difficulty to solve it for $\theta(x)$.
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