How would you approach the following equation
$$x^4 f'' - \alpha x^3 f' - \left( \beta x^2 + \gamma x - 1\right) f -1 =0$$
where $\alpha$, $\beta$, and $\gamma$ are positive constants?
The solution I need has to be finite in the origin, from which it can be seen that the conditions $f(0)=1$ and $f'(0) = \gamma$ are satisfied.
Hint:
Let $t=\dfrac{1}{x}$ ,
Then $\dfrac{df}{dx}=\dfrac{df}{dt}\dfrac{dt}{dx}=-\dfrac{1}{x^2}\dfrac{df}{dt}=-t^2\dfrac{df}{dt}$
$\dfrac{d^2f}{dx^2}=\dfrac{d}{dx}\left(-t^2\dfrac{df}{dt}\right)=\dfrac{d}{dt}\left(-t^2\dfrac{df}{dt}\right)\dfrac{dt}{dx}=\left(-t^2\dfrac{d^2f}{dt^2}-2t\dfrac{df}{dt}\right)\left(-\dfrac{1}{x^2}\right)=\left(-t^2\dfrac{d^2f}{dt^2}-2t\dfrac{df}{dt}\right)(-t^2)=t^4\dfrac{d^2f}{dt^2}+2t^3\dfrac{df}{dt}$
$\therefore\dfrac{1}{t^4}\left(t^4\dfrac{d^2f}{dt^2}+2t^3\dfrac{df}{dt}\right)-\dfrac{\alpha}{t^3}\left(-t^2\dfrac{df}{dt}\right)-\left(\dfrac{\beta}{t^2}+\dfrac{\gamma}{t}-1\right)f-1=0$
$\dfrac{d^2f}{dt^2}+\dfrac{\alpha+2}{t}\dfrac{df}{dt}+\left(1-\dfrac{\gamma}{t}-\dfrac{\beta}{t^2}\right)f-1=0$
$t^2\dfrac{d^2f}{dt^2}+(\alpha+2)t\dfrac{df}{dt}+(t^2-\gamma t-\beta)f-t^2=0$
Which relates to an ODE of the form http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=250.