Parameterized differential equation limit solution

Question

Consider the differential equation: $$x'(t)=\beta\left(-\frac{x}{1-\sin(t)/2}+\frac{2-x}{1+\sin(t)} \right) $$ I would like to prove that when $\beta\to 0$ then the solution converges to $x(t)= 2$ as suggested by numerical results.

Answer 1

As $\beta \rightarrow 0$, $x'(t) \rightarrow 0$, so $x(t) = c$ for some constant $c$, which depends on some initial condition which has to be given.

Edit:

The ODE is of the form $x'(t)+P(t)x(t)+Q(t)=0$ with $P(t) = \beta \left( \frac{4+\sin(t)}{(1+\sin(t))(2-\sin(t))}\right), Q(t) = -\beta\frac{2}{1+\sin(t)}$, which are continuous for $t \neq -\pi/2 + 2\pi k$. Therefore a unique solution must exist in the neighborhood of any initial condition, so for a finite $\beta$ we can always find a solution passing through any given point $(t_0,x_0), t_0 \neq -\pi/2 + 2\pi k$.

If your assumption is correct, then as $\beta \rightarrow 0$, and taking $x_0 \neq 2$ we should have a spike with diverging derivative at $t_0$, which means that $x(t)$ diverges, because $P(t), Q(t)$ and $\beta$ are bounded in the neighborhood of $t_0$.

Do the numerical results give spikes with diverging $x$ near $t_0$?