MHB Can This Differential Equation Be Solved with Given Initial Conditions?

[Euge]

Here's this week's problem. Happy holidays, everyone!

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Problem. Find the solution of the differential equation

$$y\ddot{y} - \dot{y}^2 - y^4 = 0$$

subject to the conditions $y(1/2) = \pi$ and $\dot{y}(1/2) = 0$.

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[Euge]

No one answered this week's problem. Here is my solution.

Spoiler

The solution is $y(t) = \pi \csc(\pi t)$. To see this, add $y^4$ to both sides of the ODE, then divide through by $y^2$ to get

$$ \frac{y\ddot{y} - \dot{y}^2}{y^2} = y^2$$

$$ \left(\frac{\dot{y}}{y}\right)^{\dot{}} = y^2$$

$$ (\log y)^{\ddot{}} = y^2$$

Employ the $u$-substitution $u = \log y$ to obtain

$$\ddot{u} = e^{2u}$$

$$\ddot{u}\, \dot{u} = e^{2u}\dot{u}$$

$$\left(\frac{\dot{u}^2}{2}\right)^{\dot{}} = \left(\frac{e^{2u}}{2}\right)^{\dot{}}$$

$$\dot{u}^2 = e^{2u} + C$$

Since $u(1/2) = \log y(1/2) = \log \pi$ and $\dot{u}(1/2) = \frac{\dot{y}(1/2)}{y(1/2)} = 0$, we deduce that $C = -\pi^2$. Thus

$$\dot{u}^2 = e^{2u} - \pi^2$$

$$\dot{u} = \pm \sqrt{e^{2u} - \pi^2}$$

$$\int \frac{du}{\sqrt{e^{2u} - \pi^2}} = \pm \int dt$$

$$\int \frac{du}{\sqrt{e^{2u} - \pi^2}} = \pm t + c$$

Letting $e^u = \pi \cosh v$, the integral on the left becomes

$$\frac{1}{\pi} \int \sech v\, dv = \frac{1}{\pi}\arccos(\sech v) + c = \frac{1}{\pi}\arccos(\pi e^{-u}) + c$$

Therefore

$$ \frac{1}{\pi} \arccos(\pi e^{-u}) = \pm t + c$$

$$ \pi e^{-u} = \cos(\pi t + c)$$

$$ e^u = \pi \sec(\pi t + c)$$

Invoking the initial condition $u(1/2) = \log \pi$, we determine $c = -\pi/2$. Since $u = \log y$, we conclude

$$y = \pi \sec\left(\pi t - \frac{\pi}{2}\right) = \pi \csc(\pi t)$$