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?

  • $\begingroup$ @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
  • 1
    $\begingroup$ @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


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.


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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