= Solution
The substitution $y=e^xg$ reduces the <homogeneous linear differential equation> to
$$
g''+\left(1-\frac1x\right)g'=0.
$$
A first <integration> gives $g'=Cxe^{-x}$, and a second gives $g=D-C(x+1)e^{-x}$. Hence
$$
\boxed{y=Ae^x+B(1+x).}
$$
The <Wronskian> of $1+x,e^x$ is $xe^x$, so they are <linearly independent> on either side of zero. At zero the differential equation has a <regular singular point>. The displayed solutions extend analytically through it; their linear combinations describe the global solutions when two continuous derivatives are required there. If no matching condition is imposed, the two sides have independent constants.
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