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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?

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  • $\begingroup$ @Luna145: The second two quantities are two ways to write the same thing, by that trig identity. $\endgroup$ Commented Jun 29, 2021 at 9:10
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    $\begingroup$ @JacobManaker Forgive me. I misunderstood the phrasing of that part. $\endgroup$
    – Moni145
    Commented Jun 29, 2021 at 9:13
  • $\begingroup$ @Luna145: That's what the comments are for! I edited into the question, so hopefully no other answerers get confused. $\endgroup$ Commented Jun 29, 2021 at 9:13
  • $\begingroup$ I have answered this question to the best of my ability. There is still a bounty on it. If you feel my answer is acceptable, feel free to accept it as the answer! $\endgroup$
    – Moni145
    Commented Jul 5, 2021 at 18:48

2 Answers 2

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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.

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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 $y_0$

enter image description here

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

enter image description here

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