ode internal name "second_order_ode_non_constant_coeff_transformation_on_B"
This method is tried to reduce the order ode the ODE by one, by doing direct transformation on \(B\left ( x\right ) \) for the ode
\[ A\left ( x\right ) y^{\prime \prime }\left ( x\right ) +B\left ( x\right ) y^{\prime }\left ( x\right ) +C\left ( x\right ) y\left ( x\right ) =0 \]
Let \[ y=Bv \] Then \(y^{\prime }=B^{\prime }v+v^{\prime }B\) and \(y^{\prime \prime }=B^{\prime \prime }v+B^{\prime }v^{\prime }+v^{\prime \prime }B+v^{\prime }B^{\prime }=v^{\prime \prime }B+2v^{\prime }B^{\prime }+B^{\prime \prime }v\) then the original ode becomes
\begin {align*} A\left ( v^{\prime \prime }B+2v^{\prime }B^{\prime }+B^{\prime \prime }v\right ) +B\left ( B^{\prime }v+v^{\prime }B\right ) +CBv & =0\\ ABv^{\prime \prime }+\left ( 2AB^{\prime }+B^{2}\right ) v^{\prime }+\left ( AB^{\prime \prime }+BB^{\prime }+CB\right ) v & =0 \end {align*}
Now we check if \(AB^{\prime \prime }+BB^{\prime }+CB=0\) or not. If it is zero, then this method works and we can now solve
\[ ABv^{\prime \prime }+\left ( 2AB^{\prime }+B^{2}\right ) v^{\prime }=0 \]
Using \(u=v^{\prime }\) which reduces the order to one.
\[ ABu^{\prime }+\left ( 2AB^{\prime }+B^{2}\right ) u=0 \]
This is first order ode now. Solved for \(u\) gives \(v^{\prime }\) which is solved for \(v\) as first order ode. Then \(y=Bv\) and we are done. This method only works of course if \(AB^{\prime \prime }+BB^{\prime }+CB=0\) comes out to be zero. Here is an example