I want to understand the solutions to ODE with removable discontinuities (conditional on specific starting values). For example, I have the function $y(x)$ defined by: \begin{align*} \sin(x) = \sin(2 y) \text{ for } x \in [0, \pi], y \in [0, 1/2 \pi]. \end{align*} It gives me the following differential equation: \begin{align*} y'(x) = \frac{\sin'(x)}{2 \sin'(2y(x))}, \quad y(0) = 0; \end{align*} where $\sin'(x) = \cos(x)$. There is a singularity at $y=\pi/2$. However, on the optimal path starting from $y(0)=0$, this singularity will be removable. Indeed, on $[0, \pi/2[$, we find that $y^*(x)= x/2$. Thus: \begin{align*} lim_{x\rightarrow\pi/2} \ y'(x) = \frac{\cos(\pi/2)}{2\cos(\pi/2) } = 1/2. \end{align*}
This is not exactly a "removable singular point", but conditional on being on this specific path, with this specific starting value, the discontinuity is not a discontinuity anymore. Is there a name for this?
Now, it seems there are two solutions to this equation (maybe I'm wrong and there are more?): \begin{align*} y_1(x) = x/2 \text{ for } x \in [0, \pi] \end{align*} or: \begin{align*} y_2(x) = \left\{ \begin{array}{ll} x/2 & \mbox{for } x \in [0, \pi/2] \\ \pi/2-x/2 & \mbox{for } x \in [\pi/2, \pi]. \\ \end{array} \right. \end{align*}
In other words, once you reach the singularity, you can either go backwards, or upwards. But I found these solutions naturally by thinking, not with equations/maths.
Moreover, I don't know how to prove that these are the only two solutions, and I'd like to prove it.
I think this property holds more generally, even in higher dimension, i.e. if I have isolated singular points, if I reach a singularity that is removable given the optimal path, I still have two (in 2D case, more in higher dimensions) solutions/directions that I can choose (at least locally) for the solution. How can I prove this? Is there any known results?
It is particularly important because it would imply that there exists a unique strictly increasing solution to these kind of problems.