Solution of the nonlinear ODE $y'' y' =A y' y + B (x-1) y$, with $y(0) = 1$, $y(1) = 0$

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What analytical techniques are available for finding solutions to the nonlinear ODE $$y'' y' =A y' y + B (x-1) y,$$ with boundary conditions $$y(0) = 1, \quad y(1) = 0,$$ where $A$ and $B$ are positive, real constants? Unfortunately, neither $A$ nor $B$ are necessarily small.

Does assuming that $y'(x) \neq 0$ allow further progress?

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Hint:

Let $t=x-1$ ,

Then $y'y''=Ayy'+Bty$ with $y(-1)=1$ and $y(0)=0$

$y''=Ay+\dfrac{Bty}{y'}$ with $y(-1)=1$ and $y(0)=0$

$\dfrac{d^2y}{dt^2}=Ay+\dfrac{Bty}{\dfrac{dy}{dt}}$ with $y(-1)=1$ and $y(0)=0$

$\dfrac{d}{dt}\left(\dfrac{dy}{dt}\right)=Ay+\dfrac{Bty}{\dfrac{dy}{dt}}$ with $y(-1)=1$ and $y(0)=0$

$\dfrac{d}{dt}\left(\dfrac{1}{\dfrac{dt}{dy}}\right)=Ay+Bty\dfrac{dt}{dy}$ with $t(1)=-1$ and $t(0)=0$

$\dfrac{d}{dy}\left(\dfrac{1}{\dfrac{dt}{dy}}\right)\dfrac{dy}{dt}=Ay+Bty\dfrac{dt}{dy}$ with $t(1)=-1$ and $t(0)=0$

$-\dfrac{\dfrac{d^2t}{dy^2}}{\left(\dfrac{dt}{dy}\right)^3}=Ay+Bty\dfrac{dt}{dy}$ with $t(1)=-1$ and $t(0)=0$

$\dfrac{d^2t}{dy^2}=-Ay\left(\dfrac{dt}{dy}\right)^3-Bty\left(\dfrac{dt}{dy}\right)^4$ with $t(1)=-1$ and $t(0)=0$

You can consider as the ODE of the type http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=429