How to solve differential equation $(y'/y)'=a((b/x^2)+xy)$?

Three types of solution:

(1)=closed form (generally)

(2)=series expansion around $0$ (special cases)

(3)=series expansion around $1$ (generally)

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$(1)$

$\displaystyle (\ln y)''=\frac{ab}{x^2}+axy$

$\displaystyle y:=cx^{-ad}$ 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) $y_1^Ay_2^B\neq Ay_1+By_2$ .

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$(2)$

$\displaystyle (\ln y)''=\frac{ab}{x^2}+axy\enspace$ 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 $\enspace b_{n,k}\in\mathbb{N}$ , $\enspace b_{n,1}=1$

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$(3)$

$\displaystyle (\ln y)''=\frac{ab}{x^2}+axy\enspace$ 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 $1$: $\enspace\displaystyle z=\sum\limits_{k=0}^\infty\frac{(x-1)^k}{k!}z^{(k)}(1)$