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
 A: Let we set $f=y'-y$. Then:
$$(x-1) y'' - x y' + y = 0 $$
gives:
$$ (x-1) f' - f = 0 $$
that is a separable ODE, solved by $f(x)=K(x-1)$. After that, we just have to solve:
$$ y'-y = K(x-1) $$
leading to:
$$ y = C e^x - Kx.$$
A: Let $y=\sum_{n=1}^{\infty}a_nx^n$.  Then, we obtain after some straightforward analysis
$$\sum_{n=0}^{\infty}\left((n+2)(n+1)a_{n+2}-n(n+1)a_{n+1}+(n-1)a_n\right)x^n=0$$

CASE 1:
We can easily show that for $a_0\ne 0$, $a_n=\frac{1}{n!}$ for $n\ge 2$.  We are free to choose $a_1$ here and choose $a_1=a_0$. 

CASE 2:
For $a_0=0$, and $a_1\ne 0$, we can easily show that $a_n=0$ for $n\ne 1$.

Thus, the two homogeneous solutions are 
$$\bbox[5px,border:2px solid #C0A000]{y_1(x)= a_0\sum_{n=0}^{\infty}\frac{x^n}{n!}=a_0e^x}$$
and 
$$\bbox[5px,border:2px solid #C0A000]{y_2(x)=a_1x}$$
A: Hint: It is easily seen that $y=x$ is a solution. Hence write $y=xu$ then the equation will be reduce to a separable first order differential equation giving you the second solution.
A: 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 $1$.
