Cauchy-Euler-like equation Is anybody able to solve the equation: $x(x-1)y''+(3x-1)y'+y=1$ without advanced techniques like series solution, nor integral tranform ? 
I learned from Wolframalpha a first step of the solution is: $\tfrac{\text d}{\text dx}(3x-1) - \tfrac{\text d^2}{\text dx^2}[x(x-1)] = 1$, and then replacing we get 
$x(x-1)y"+(3x-1)y'+\left(\tfrac{\text d}{\text dx}(3x-1) - \tfrac{\text d^2}{\text dx^2}[x(x-1)]\right)y=1$. But I have no idea then and I don't have a pro account for wolframalpha. So what is the next step ?
 A: My solution (at least for my opinion) is very messy and I suppose there is a lot cleaner and easier solution. I hope it will give you some direction.
Expanding your line we have $$x\frac{\text{d}^2y}{\text{d}x^2}(x-1)+\frac{\text{d}y}{\text{d}x}(3x-1)+\frac{\text{d}}{\text{d}x}(3x-1)y-\frac{\text{d}^2}{\text{d}x^2}(x(x-1))y=1$$Adding and substract $\displaystyle \frac{\text{d}y}{\text{d}x}\frac{\text{d}}{\text{d}x}(x(x-1))$ we have\begin{align*}&\left[\frac{\text{d}y}{\text{d}x}\frac{\text{d}}{\text{d}x}(x(x-1))+x\frac{\text{d}^2y}{\text{d}x^2}(x-1)\right]+\left[\frac{\text{d}y}{\text{d}x}(3x-1)+\frac{\text{d}}{\text{d}x}(3x-1)y\right]\\-&\left[\frac{\text{d}y}{\text{d}x}\frac{\text{d}}{\text{d}x}(x(x-1))+\frac{\text{d}^2}{\text{d}x^2}(x(x-1))y\right]=1\end{align*}Now, using the product rule we have$$\frac{\text{d}}{\text{d}x}\left(x\frac{\text{d}y}{\text{d}x}(x-1)\right)+\frac{\text{d}}{\text{d}x}((3x-1)y)-\frac{\text{d}}{\text{d}x}((2x-1)y)=1$$And eventually$$\frac{\text{d}}{\text{d}x}\left(x\frac{\text{d}y}{\text{d}x}(x-1)+xy\right)=1$$Integrating both sides with respect to $x$ $$x\frac{\text{d}y}{\text{d}x}(x-1)+xy=x+C$$Dividing by $x(x-1)$ we get $\displaystyle y'+\frac{y}{x-1}=\frac{x+C}{x(x-1)}$. We have $x-1$ as the integrating factor, thus$$(x-1)y'+y=\frac{x+C}{x}\Rightarrow \frac{\text{d}}{\text{d}x}((x-1)y)=1+\frac{C}{x}$$
Integrate with respect to $x$, divide by $x-1$ and you will get the answer.
