# Solving $xy''-(1-x)y'+y=0$

$$xy''-(1-x)y'+y=0$$

So I know how to solve this via power series. Recently, a friend of mine was asking me how one could solve this without using series. I've got no real idea how to answer this without blinding guessing one of the solutions. Anyone have ideas on how to approach without series solution?

$$xy''-(1-x)y'+y=0$$ expanded $$xy'' -y' +xy' + y = xy'' -y' + \frac{d}{dx}xy$$ now using the fact $$\frac{d}{dx}xy' = xy'' + y'$$ and manipulating to yield $$\frac{d}{dx}xy' - 2y' = xy'' - y'$$ we can rewrite the original equation as
$$xy'' -y' +xy' + y = \frac{d}{dx}xy' - 2y'+ \frac{d}{dx}xy = \frac{d}{dx}\left[xy' - 2y + xy\right]=0$$ I believe you can take it from there right? :)