Solution
= Solution
Substitution of $y=e^{-x}$ gives
$$
(x+\lambda+1)e^{-x}-(x+\lambda)e^{-x}-e^{-x}=0.
$$
For a second solution $y=ax+b$, the <ordinary differential equation> reduces to $a\lambda-b=0$. Taking $a=1$ gives the linearly independent solution
$$
\boxed{y_1=x+\lambda}.
$$