Let us be given the equation Ayy' + By'' = 0.
Determine \color{red}a \neq 0 such that t \mapsto y(t) = \dfrac{\color{red}a}{t} is a solution.
\color{red}a = 2*B/A The idea is to use {\color{blue}y(t)= \dfrac{a}{t}} with {\color{blue}y'(t)} and {\color{blue}y''(t)} in the ODE Ayy' + By'' = 0.
The derivatives are {\color{blue}y'(t) = -a t^{-2} = -\dfrac a {t^2}} and {\color{blue}y''(t) = 2a t^{-3} = 2 \dfrac a {t^3}}
and thus we get A \cdot {\color{blue}\dfrac{a}{t}} \cdot \left({\color{blue}-\dfrac a {t^2}}\right) +negParens(B)\cdot {\color{blue} 2 \dfrac a {t^3}}= 0
Upon sorting terms we arrive at
\dfrac{-A{\color{red}a}^2 + 2*B{\color{red}a}} {t^3}= 0.
This equation is fulfilled for all t (with t \neq 0) if the numerator -A{\color{red}a}^2 + 2*B{\color{red}a}= 0.
\color{red}a, to find {\color{red}a} = fractionReduce(2*B,A) (or {\color{red}a} =0).