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 $(-\infty,\infty)$.
 A: Your equation has the form
$$
-2f'(2θ)θ'^2-f(2θ)θ''=0,~~f(2θ)=\cos(2θ)+c\sin(2θ)
$$
This is singular for $f(2θ)=0$, that is, the lines $ θ=k\pi-\arctan\frac1c$, $k\in\Bbb Z$.
Divide by $θ'f(2θ)$ and integrate, or integrate directly using the product and chain rule, to get
$$
f(2θ)θ'=C.
$$
Inserting $f(2θ)=\cos(2θ)+c\sin(2θ)$ and integrating again leads to
$$
\sin(2θ)-c\cos(2θ)=2Cx+D
$$
Obviously, the range on the left side is limited, while the range on the right side is unlimited. However, local solutions exist for all $c$. With $c=\tan\phi$ the equation can be further transformed to
$$
\sin(2θ-ϕ)=Ax+B\implies θ=\frac12(ϕ+\arcsin(Ax+B))
$$
with constants $A,B$ determined by the initial conditions. It might be necessary to correct $\phi$ by multiples of $\pi$ to reach the initial point.
Solutions approach the singular lines with a vertical tangent. While it is possible to patch solutions together continuously, there is no natural or unique continuation.
A: $$2 (\frac{d \theta}{dx})^2 [\sin(2\theta) -c \cos(2 \theta)]-[\cos(2\theta) +c \sin(2 \theta)]\frac{d^2}{dx^2}\theta=0$$
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 $\theta(x)$.
A: The given equation can be easily integrated:
$$-\dfrac12\left((\cos 2\theta+c\sin2\theta)\theta'\right)'=0\Rightarrow
(\cos 2\theta+c\sin2\theta)\theta' = \dfrac12C_1,\\
(\cos 2\theta+c\sin2\theta)\,\mathrm d(2\theta) = C_1\,\mathrm dx\Rightarrow 
\sin 2\theta-c\cos2\theta = C_1 x+C_2 = y,$$
wherein the constants $C_1,C_2\ $ can be defined from the initial conditions.
Let
$$t = \tan\theta\Rightarrow \cos2\theta = \dfrac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta} = \dfrac{1-t^2}{1+t^2},\quad \sin2\theta = \dfrac{2\sin\theta\cos^2\theta}{\cos^2\theta+\sin^2\theta} = \dfrac{2t}{1+t^2},$$
$$2t-c(1-t^2) = (1+t^2)y,\quad (c-y)t^2 + 2t - (c+y) = 0,$$
with the common solutions
$$\small t_{1,2} = \dfrac{1\pm\sqrt{1+c^2-y^2}}{c-y},$$
$$\cos2\theta = \dfrac{-cy\mp\sqrt{1+c^2-y^2}}{c^2+1},\ \sin2\theta = \dfrac{-y\pm c\sqrt{1+c^2-y^2}}{c^2+1}, \tag1$$
and the special case
$$y=c\Rightarrow t=c,\quad \cos2\theta =\dfrac{1-c^2}{1+c^2},\quad \sin2\theta =\dfrac{2c}{1+c^2}.\tag2$$
If $y^2 > 1+c^2,\ $ then the solutions $(1)$ are complex. This fact points to the bounded values of the variable $x.$
