Find a general solution of the following equation:
$x' \sin(2t)=2(x+\cos(t))$.
Determine a solution limited by $t \to \frac{\pi}{2} $.
$x' = \frac{2(x+\cos(t))}{\sin(2t)} \Rightarrow x' - \frac{2x}{\sin(2t)} = \frac{\cos(t)}{\sin(2t)} $
First we can solve a homogeneous equation, and then after variation of parameters we can solve inhomogeneous equation. That's right, but there will be very difficult integers to compute, and even if I will be able to solve it (after some time), I don't know what to do with "Determine a solution limited by $t \to \frac{\pi}{2} $".
I think that there may be some different, easier way to solve our equation, but yet, I don't recognize any. Can we transform this to exact differential equation?
Btw wolfram alpha says that the solution is $x(t)=C_1 tan(t)-sec(t)$, don't know if it will help but... .
I will be grateful for any help.
This is an answer to your second question only.
Assume that we know that $$ \frac{C \sin{t} - 1}{\cos{t}} $$ is a general solution of the ODE. Of course, for no value of $C$ the above function is defined at $t = \pi/2$. But we want to find $C$ such that the function has a (finite) limit as $t \to \pi/2$.
The denominator tends to zero as $t \to \pi/2$, so a necessary condition is that $C \sin{t} - 1 \to 0$ as $t \to \pi/2$. The only possibility left is $C = 1$.
Notice that this is only a necessary condition. We have to check that $$ \frac{\sin{t} - 1}{\cos{t}} $$ tends to a finite limit as $t \to \pi/2$. Indeed, this limit is $0$.