Answer
$y (x)= x e^x \in ker (L)$
Work Step by Step
We have: $Ly=(-D^2+2D-1)(x e^x)$
$L(x e^x)=(-2 e^x+x e^x)+2(e^ x+x e^x)-x e^x\\=-2e^x-x e^x+2 e^x+2x e^x-x e^x \\=0$
This yields $y (x)= x e^x \in ker (L)$
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