Solving the second-order nonlinear o.d.e. $y(y-xy')+(a^2-x^2)yy^{"}=b^2$ arising from a metrizability problem When we was checking the properties of metrizability of specific geometric spaces we encountered the following second-order nonlinear DE:
$$y(y-xy')+(a^2-x^2)yy^{"}=b^2$$
where $y$ depends on $x$ and $a$ and $b$ are nonzero real constants.
We have have tried to apply routine techniques, but they didn't work!
 A: As far as I can see there isn't a nice closed form for general $a, b$, but notice that the change of variables $x = a \sin \theta, y = b u$ yields the simpler, autonomous equation
$$u (u'' + u) = 1.$$ This form makes more apparent the constant solutions $u = \pm 1$, i.e., $y = \pm b$.
If we now treat $\theta$ as a function of $u$, our differential equation transforms to
$$u \left[-\frac{\theta''}{(\theta')^3} + u\right] = 1 ,$$
which is a separable first-order equation in $\theta'$. Separating and integrating gives
$$(\theta')^{-2} = C + 2 \log |u| - u^2,$$
solving for $\theta'$ and integrating again gives
$$\theta = \pm \int_1^u \frac{dt}{\sqrt{C + 2 \log |t| - t^2}} + D ,$$
and finally reversing the substitution yields
$$\boxed{\sin \frac{x}{a} = \pm \int_b^y \frac{d\tau}{\sqrt{C + 2 b^2 \log |\tau| - \tau^2}} + D} .$$
The arbitrary constant $C$ in this final expression is different from the constant with that name in the two previous equations.
Remark The limiting case $b = 0$ is somewhat interesting, too:
If $a \neq 0$, making just the change $x = a \sin \theta$ gives
$$y(\theta) (y''(\theta) + y(\theta)) = 0,$$
which has apparent general solution
$$y(\theta) = C \sin \theta + D \cos \theta = \hat{C} \sqrt{a^2 - x^2} + \hat{D} x .$$ On the other hand, if $a = 0$, the equation simplifies to
$$x^2 y'' + x y' - y = 0,$$
after which the ansatz $y(x) = x^m$ yields $m = \pm 1$, and so
$$y = C x + \frac{D}{x} .$$
