A trigonometric non-linear differential equation I am stuck with the following initial value problem:
\begin{equation}
\dot{Y}_t = 4 \arctan(1/Y_t) = 2\pi - 4\arctan(Y_t), \quad Y_0 = y > 0
\end{equation}  (These are the same thing, by a trig identity.)
I tried rewriting a couple of times, but didn't succeed in determining an explicit solution. Is there one? I appreciate any hint.
Edit: If we differentiate, the above line, we get
$$
(1+Y_t^2)\ddot{Y}_t + 4 \dot{Y}_t = 0.
$$
This seems helpful. Is this explicitly solvable?
 A: Let $v(y) = y'$, and our differential equation transforms into:
$$v'\cdot v(1+y^2)+4v=0$$
$$\implies v(v'(1+y^2) + 4)=0$$
So, $v=0$ or $ v'(1+y^2)+4=0$. $v=0$ gives the solution $v=c$, but this is not interesting. Suppose $v'(1+y^2) + 4 = 0$. Straight integration yields $v(y) = -4\arctan(y) + c_1$. Back-substitute $v(y)= y'$.
If $y' = -4\arctan(y)+c_1$, then separation of variables yields
$$
\frac{dy}{-4\arctan(y) + c_1} = dx \iff x =  \int_{0}^{y} \frac{du}{-4\arctan(u) + c_1}.
$$
I couldn't prove it to you, but from doing lots and lots of integrals in my day, I can tell you with pretty much certainty by looking at it that $\int_{0}^{y} \frac{du}{-4\arctan(u) + c_1}$ doesn't have a nice closed form. And, even then, if you wanted it explicitly, you'd have to solve for $y$. So, short answer: no, it's not explicitly solvable. But if you're able to use numerical techniques, this is probably the way to do it.
A: Not likely exactly solvable, as mentioned in Luna145's answer. Fortunately, it is straightforward to find the leading behavior for large $t$.
With the given initial condition $y_0>0$, we have that $(2\pi-4\tan^{-1}y)>0$ for all $t$, so $y$ is an increasing function. We expand around large $y$ which corresponds to large $t$
$$
y'=2 \pi -4\tan^{-1}(y) \sim 4y^{-1} \qquad ;\qquad y \to \infty
$$
We are left with the much simpler equation
$$
yy'=4
$$
The solution is
$$
y(t)=\sqrt{8t+y_0^2}
$$
Here is a plot of the numerical solution and the approximation for various values of $y_0$

And a plot of the ratio $y_{\text{numeric}}/\sqrt{8t+y_0^2}$

