The equation $yy''=-1$ Is the equation
$$yy''=-1$$
well-known in literature, i.e., are there any references where it has been studied?
It came up in the context of my research and, due to its simple appearance, I thought that the behaviour of its solutions might be known already.
 A: Solution involving an integral
Around $x_0$ and supposing $y(x_0) \neq 0$ you have
$$y^{\prime \prime} y^\prime= - \frac{y^\prime}{y}$$ and therefore
if $y_0,y_0^\prime$ denotes $y(x_0),y^\prime(x_0)$ respectively
$$\frac{1}{2} \left(y^\prime\right(x))^2 -\frac{1}{2} \left(y^\prime_0\right)^2= - \ln \left\vert \frac{y(x)}{y_0} \right\vert$$
From there
$$\int_{y_0}^{y(x)} \frac{dt}{\sqrt{\left(y^\prime_0\right)^2 -2\ln \left\vert \frac{t}{y_0} \right\vert}} = \pm(x-x_0).$$ Which provides a way to study the solutions based on the study of the map
$$y \mapsto \int_{y_0}^{y} \frac{dt}{\sqrt{\left(y^\prime_0\right)^2 -2\ln \left\vert \frac{t}{y_0} \right\vert}}.$$
A: $$yy''=-1$$
Substitute $p=\dfrac {dy}{dx}$:
$$y\dfrac {dp}{dx}=-1$$
$$y\dfrac {dp}{dy}\dfrac {dy}{dx}=-1$$
$$py\dfrac {dp}{dy}=-1$$
This is a seprable DE.
A: Switch variables to make
$$\frac{x''}{[x']^3} y=1$$ Reduction of order $$p=x' \implies \frac{p'}{p^3} y=1\implies p=\pm \frac 1{\sqrt{c_1-2\log(y)}}$$ which gives
$$x+c_2=\pm \sqrt{\frac{\pi\, e^{c_1}}{2}} \, \text{erf}\left(\frac{\sqrt{c_1-2 \log
   (y)}}{\sqrt{2}}\right)$$
A: Here are some thoughts on the equation, which suggest an iterative approach for approximating solutions.
Via scaling and shifting, any solution (corresponding to any legal pair of initial conditions; in particular when $y_0 \neq 0$) may be realized from a solution to the IVP
$$\left\{
\begin{array}{l}
y(x)y''(x) = -\lambda^2\\
y(0) = 1\\
y'(0) = m
\end{array}
\right.$$
where $\lambda \neq 0$ and $m$ are real constants, and the solution $y$ is defined upon a maximal interval $\mathcal I = (a,b) \ni 0$. By continuity and the fact that $y(0) = 1$ one observes $y(x) > 0$ for each $x \in \mathcal I$, which further implies $y''(x) < 0$ for each $x$. This provides insight as to why no solution can be defined for all real $x$; the solution must always be positive while always curving downward, hence it collides with the $x$-axis in a finite time.
An easy upper bound for $y$ is $f_0(x) = 1 + mx$, but we shall show that from any upper bound a better one may be constructed. Imagine an object of mass $1$ falling with trajectory given by $y(x)$; it experiences a downward force with a (positive) magnitude $F(x) = -y''(x)$. Then for $x > 0$, the bound $y(x) \leq f_0(x)$ yields
$$F(x) = \frac{\lambda^2}{y(x)} \geq \frac{\lambda^2}{f_0(x)} \implies y''(x) \leq -\frac{\lambda^2}{f_0(x)}$$
Integrating twice through the bound and applying FTOC yields
$$y(x) \leq 1 + mx -\lambda^2 \int_0^x\int_0^u \frac1{f_0(v)} \,dv \,du$$
and finally changing the order of integration (and evaluating the inner integral) provides a new upper bound
$$y(x) \leq f_1(x) := 1 + mx - \lambda^2 \int_0^x \frac{x - v}{f_0(v)}\, dv$$
Additional computation may be needed for the case when $x < 0$. The above construction can be iterated to produce a sequence of functions $(f_n)_n$ which all bound $y(x)$ from above, and numerical experimentation suggests that the sequence is decreasing.

This also suggests that the sequence has a limit $f(x) = \lim_n f_n(x)$, and that the limit satisfies
$$f(x) = 1 + mx -\lambda^2 \int_0^x \frac{x-v}{f(v)}\, dv$$
It is possible to prove that in this case $f$ satisfies the same IVP as $y$, and so by uniqueness we have $y(x) = f(x) = \lim_n f_n(x)$ for each $x \in \mathcal I$.
From $f_0(x) = 1 + mx$, it is possible (though tedious) to compute $f_1(x)$ in closed form, but perhaps not $f_2$. It is important to note that $f_1(x)$ has a root (this may be verified by hand), which imposes a bound on the interval of existence $\mathcal I$. In the special case where $m = 0$ and $\lambda^2 = 1$, I have calculated
$$\begin{align*}
f_0(x) &= 1\\
f_1(x) &= 1 - \frac12 x^2\\
f_2(x) &= 1+\frac{1}{\sqrt{2}}x\ln\left(\frac{\sqrt{2}-x}{\sqrt{2}+x}\right)+\ln\left(\frac{2}{2-x^{2}}\right)
\end{align*}$$
Here the three bounds are plotted against the true solution $\operatorname{erf}(\sqrt{-\ln y}) = \sqrt{2/\pi}|x|$

