Consider the Cauchy problem defined by
$$ \left\{ \begin{array}{c} y' = e^{-x^2}\sin(y) \\ y(0) = a \end{array} \right. $$
with $ a \in \mathbb R $.
If $ y_a(x) $ is the solution, for which values of $a$ it's true that $ \pi < y_a(x) < 2\pi$?
Prove that for $ a = -\pi/2 $ the solution is defined on the whole real line and it's strictly decreasing.
It seems to me that to be able to answer the questions one must find $y_a(x)$ explicitly. However, the only way i could think of solving the differential equation is by variable separation. This leads to the infamous integral of $e^{-x^2}$, that I'm only able to solve by going to two variable and using polar coordinates. The problem is getting back to x and y needs you to limit the domain of x and y, making things more difficult than I believe they should be given the time requisite of about 30 minutes. I suppose that the way intended by the professor to solve this problem is to search for some sort of indirect solution in which $y_a(x)$ is not explicited.
Does someone have any suggestions? Thanks
In order to answer this question, it should be noted that the points $y=\pi k$, $k\in\mathbb Z$, are equilibrium points. This means that all other solutions (by the existence and uniqueness theorem) cannot take these values. But since the solutions are continuous, this implies that each of them can only take values from one of the intervals $I_k=(\pi k, \pi(k+1))$:
So the answer to the first question is $a\in (\pi,2\pi)$.
To answer the second question, it must be said that since the right side of the equation satisfies everywhere the condition of the existence and uniqueness theorem, the only way for the solution to cease to exist is to blow up to infinity. But then it is necessary to cross one of the "red lines", i.e. through the equilibrium position. Again, this is impossible by the same theorem.
To prove that the solution is strictly decreasing, one needs to examine the sign of the right side of the equation in the corresponding interval.