How to solve differential equation $(y'/y)'=a((b/x^2)+xy)$? In a physics problem I'm working on, the following differential equation has appeared
$$\left(\frac{y'}{y}\right)'=a\left(\frac{b}{x^2}+xy\right),~~(x>0)$$
The first approaches tried were those that were outlined in this previous question of mine. Alas, there is a new complicating term (the $1/x^2$ one) that makes this quite complicated.
This time, wolframalpha doesn't give me a closed form solution, so I don't know exactly what to expect.
If it helps, I'm looking for solutions that are big but not divergent near the origin (i.e. for small values of $x$), and that decay down (along with all its derivatives) to 0 at infinity..
 A: Three types of solution:
(1)=closed form (generally) 
(2)=series expansion around $0$ (special cases)
(3)=series expansion around $1$ (generally) 

$(1)$
$\displaystyle (\ln y)''=\frac{ab}{x^2}+axy$
$\displaystyle y:=cx^{-ad}$  where $c$ and $d$ unknown and $a\ne 0$
$\displaystyle \frac{ad}{x^2}=\frac{ab}{x^2}+\frac{ac}{x^{ad-1}}$
=>  $\enspace\displaystyle d=\frac{3}{a}\enspace$ and $\enspace\displaystyle  c=\frac{3}{a}-b$
One solution is $\enspace\displaystyle y:=(\frac{3}{a}-b)x^{-3} \,$ . 
A second solution would be linear independend of $y$.  
Note:
The homogeneous ($b=0$) and inhomogeneous ($b\neq 0$) solution have the same term here. 
It’s not possible to put two solutions $y_1$ and $y_2$ together to one solution $y:=Ay_1+By_2$ ($A,B\in\mathbb{R}$), because it’s (generally) $y_1^Ay_2^B\neq Ay_1+By_2$ . 

$(2)$ 
$\displaystyle (\ln y)''=\frac{ab}{x^2}+axy\enspace$ with $\enspace y:=x^{-ab}e^z$ 
It follows $z''=ax^{1-ab}e^z$ . 
A series expansion around $0$ is only possible for $1-ab\in\mathbb{N}_0$.
$(a)\enspace$ I show the expansion here for the homogeneous case $b=0$ : $\enspace y:=e^z$.
$z''=axe^z$ 
A Taylor series expansion around $0$ for $z$ can be created by:
$z(0):=C_1$ 
$z'(0):=C_2$
$z''(0)=0$ 
$z'''(0)=ae^{C_1}$
$z^{(4)}(0)=2C_2ae^{C_1}$
$z^{(5)}(0)=3C_2^2ae^{C_1}$
$z^{(6)}(0)=4C_2^3ae^{C_1}+4(ae^{C_1})^2$
$z^{(7)}(0)=5C_2^4ae^{C_1}+30C_2(ae^{C_1})^2$
$z^{(8)}(0)=6C_2^5ae^{C_1}+138C_2^2(ae^{C_1})^2$
$z^{(9)}(0)=7C_2^6ae^{C_1}+504C_2^3(ae^{C_1})^2+98(ae^{C_1})^3$
$z^{(10)}(0)=8C_2^7ae^{C_1}+1160C_2^4(ae^{C_1})^2+800C_2(ae^{C_1})^3$
...
$n\geq 2$ : $\enspace\displaystyle z^{(n)}(0)=\sum\limits_{k=1}^{\lfloor\frac{n}{3}\rfloor}a_{n,k}C_2^{n-3k}(ae^{C_1})^k\enspace$ with $\enspace a_{n,k}\in\mathbb{N}$ , $\enspace a_{n,1}=n$
Series around $0$: $\enspace\displaystyle z=\sum\limits_{k=0}^\infty\frac{x^k}{k!}z^{(k)}(0)$
$(b)\enspace$ But the easiest case here is $ab=1$ with $\enspace y:=e^z$.
$z''=ae^z$  
$z(0):=C_1$ 
$z'(0):=C_2$
The coefficients of a Taylor series expansion around $0$ for $z$ with $n\geq 2$ are 
$\enspace\displaystyle z^{(n)}(0)=\sum\limits_{k=1}^{\lfloor\frac{n}{2}\rfloor}b_{n,k}C_2^{n-2k}(ae^{C_1})^k\enspace$ with $\enspace b_{n,k}\in\mathbb{N}$ , $\enspace b_{n,1}=1$

$(3)$ 
$\displaystyle (\ln y)''=\frac{ab}{x^2}+axy\enspace$ with $\enspace y:=x^{-ab}e^z$ 
It follows $z''=ax^{1-ab}e^z$ .
A series expansion around $1$ for $z$ begins with :
$z(1):=D_1$ 
$z'(1):=D_2$
$z''(1)=ae^{D_1}$ 
$z'''(0)=-ae^{D_1}(ab-1-D_2)$
$z^{(4)}(0)=ae^{D_1}((ab-1)ab-2(ab-1)D_2+D_2^2)+(ae^{D_1})^2$
$z^{(5)}(0)=-ae^{D_1}((ab+1)(ab-1)ab-3ab(ab-1)D_2+3(ab-1)D_2^2-D_2^3)$
$\enspace \enspace \enspace \enspace \enspace \enspace -(ae^{D_1})^2(4(ab-1)-4D_2)$
... 
Series around $1$: $\enspace\displaystyle z=\sum\limits_{k=0}^\infty\frac{(x-1)^k}{k!}z^{(k)}(1)$
A: Hint:
Let $y=x^{-ab}e^u$ ,
Then $y'=x^{-ab}e^uu'-abx^{-ab-1}e^u=x^{-ab}e^u(u'-abx^{-1})$
$\therefore(u'-abx^{-1})'=\dfrac{ab}{x^2}+ax^{1-ab}e^u$
$u''+\dfrac{ab}{x^2}=\dfrac{ab}{x^2}+ax^{1-ab}e^u$
$u''=ax^{1-ab}e^u$
Approach $1$:
Follow the method in http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=440:
$\therefore\dfrac{d^2x}{du^2}=-ae^ux^{1-ab}\left(\dfrac{dx}{du}\right)^3$
When $ab\neq2$ ,
Let $\begin{cases}t=\dfrac{dx}{du}\\v=x^{2-ab}\end{cases}$ ,
Then $\dfrac{d^2v}{dt^2}=\dfrac{tv^{-\frac{ab-1}{ab-2}}}{ab-2}\left(\dfrac{dv}{dt}\right)^2$
Which reduces to a generalized Emden–Fowler equation.
Approach $2$:
Follow the method in http://eqworld.ipmnet.ru/en/solutions/ode/ode0314.pdf:
Let $\begin{cases}t=x^{3-ab}e^u\\v=xu'\end{cases}$ ,
Then $t(v+3-ab)v'=at+v$
Which reduces to an Abel equation of the second kind.
