I am trying to solve the equation $$\frac{y''}{y'}=\frac{\alpha}{r y^2}$$, which is of the form of the Emden–Fowler equation.
I found a solution in chapter 2.5.2.5 of Handbook of Exact Solutions for Ordinary Differential Equations, Chapman & Hall/CRC, however i don't understand how to derive it. I need this understanding because i get this equation from differentiating another one, and so i have to impose a constrain on the initial conditions.
The solution is apparently given by, $$ r =\pm \exp\left [\left (\int\text{d}\tau \tau^{-3/2}(2\tau^{1/2} - 2 \tau^{-1/2} +C_1)^{-1}\right) +C_2\right ]\,, \tau=\frac{1}{\alpha y^2}$$ So my question is how to derive this result from the equation. Help much appreciated.
[UPDATE: I solved] I simply noticed that, $$r y'' = \alpha \left(-\frac{1}{y}\right )'\implies \int\text{d}r \,r y'' = (ry' - y)\Big|_{y_0,r_0}^{y,r} = -\frac{1}{y}\Big|_{y_0}^y$$ integrating once more yields to the desired result.