<s>[INST] Hints on solving $y''-\frac{x}{x-1}y'+\frac{1}{x-1}y=0$ 
$$y''-\frac{x}{x-1}y'+\frac{1}{x-1}y=0$$

Is there any simple method to solve this equation? 
I need hints please $\color{red}{not}$ a full answer
 [/INST] Applying the idea in my comment:
$$0 = y'' - \frac{x-1+1}{x-1}y' + \frac{1}{x-1}y = y'' -\frac{1}{x-1}y' - y' + \frac{1}{x-1}y.$$
Multiplying by $\dfrac{1}{x-1}$, this becomes
$$ 0 = \frac{1}{x-1}y'' - \frac{1}{(x-1)^2}y' - \frac{1}{x-1}y' + \frac{1}{(x-1)^2}y.$$
We can realize the first two terms together as $\dfrac{d}{dx}\left(\dfrac{1}{x-1}y'\right)$ and the second two terms can similarly be realized as $-\dfrac{d}{dx}\left(\dfrac{1}{x-1}y\right).$ (This is just the method of integrating factors in disguise.) Putting these two together gives
$$ 0 = \frac{d}{dx}\left(\frac{1}{x-1}y'\right) - \frac{d}{dx}\left(\frac{1}{x-1}y\right) = \frac{d}{dx}\left(\frac{1}{x-1}(y'-y)\right).$$
That is to say that
$$ C = \frac{1}{x-1}(y'-y).$$
This can be solved by integrating factors as well. This method has the benefit of not doing a perhaps strange change of function but it does require a clever form of $1$.
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