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
3.3.2.12.1 Example 1 \[ xy^{\prime \prime }+\left ( -1-x\right ) y^{\prime }+y=0 \]
Here \(A=x,B=\left ( -1-x\right ) \) and \(C=1\), hence \(B^{\prime }=-1,B^{\prime \prime }=0\) and therefore
\begin {align*} AB^{\prime \prime }+BB^{\prime }+CB & =0+\left ( -1-x\right ) \left ( -1\right ) +\left ( -1-x\right ) \\ & =1+x-1-x\\ & =0 \end {align*}
It works. Hence the reduces ode becomes
\[ ABv^{\prime \prime }+\left ( 2AB^{\prime }+B^{2}\right ) v^{\prime }=0 \]
Let \(u=v^{\prime }\) then
\begin {align*} ABu^{\prime }+\left ( 2AB^{\prime }+B^{2}\right ) u & =0\\ x\left ( \left ( -1-x\right ) \right ) u^{\prime }+\left ( -2x+\left ( -1-x\right ) ^{2}\right ) u & =0\\ u-xu^{\prime }+ux^{2}-x^{2}u^{\prime } & =0\\ u^{\prime }\left ( -x-x^{2}\right ) +u\left ( 1+x^{2}\right ) & =0\\ u^{\prime }-\frac {\left ( 1+x^{2}\right ) }{\left ( x+x^{2}\right ) }u & =0 \end {align*}
This is linear first order ode solved using integrating factor which gives
\[ u=c_{1}\frac {xe^{x}}{\left ( 1+x\right ) ^{2}}\]
Hence since \(v^{\prime }=u\) then
\[ v^{\prime }=c_{1}\frac {xe^{x}}{\left ( 1+x\right ) ^{2}}\]
This is quadrature. Solving gives
\[ v=c_{2}+c_{1}\frac {e^{x}}{1+x}\]
Therefore
\begin {align*} y & =Bv\\ & =\left ( -1-x\right ) \left ( c_{2}+c_{1}\frac {e^{x}}{1+x}\right ) \\ & =c_{2}\left ( 1+x\right ) +c_{1}e^{x} \end {align*}
Note that this method is sensitive to the ODE is written. If we divide the ode by \(A\) is becomes
\[ y^{\prime \prime }+\frac {\left ( -1-x\right ) }{x}+\frac {1}{x}y=0 \]
And now \(A=1,B=\frac {\left ( -1-x\right ) }{x}\) and \(C=\frac {1}{x}\), hence \(B^{\prime }=-\frac {1}{x}+\frac {1+x}{x^{2}}\) and \(B^{\prime \prime }=\frac {2}{x^{2}}-\frac {2}{x^{3}}\left ( 1+x\right ) \) then
\begin {align*} AB^{\prime \prime }+BB^{\prime }+CB & =\left ( \frac {2}{x^{2}}-\frac {2}{x^{3}}\left ( 1+x\right ) \right ) +\left ( \frac {\left ( -1-x\right ) }{x}\right ) \left ( -\frac {1}{x}+\frac {1+x}{x^{2}}\right ) +\frac {1}{x}\left ( \frac {\left ( -1-x\right ) }{x}\right ) \\ & =-\frac {1}{x^{3}}\left ( x^{2}+2x+3\right ) \\ & \neq 0 \end {align*}
So this method now fails to reduce the ode order by one. So in practice, I try first on the ode as given, and then try again by normalizing it so that \(B\) is not rational function and try again. In other words, given an ode \(y^{\prime \prime }+\frac {\left ( -1-x\right ) }{x}+\frac {1}{x}y=0\,\) then try with \(B=\frac {\left ( -1-x\right ) }{x}\) and if this fails, try again after multiplying the ode by \(x\) so now \(B=\left ( -1-x\right ) \) and \(A=x\) and \(C=1\) and see if this works or not. This method of course only works when \(B\) is not zero.