= Solution
Let $D=d/dx$ and write the constant-coefficient <linear ordinary differential equation> as $p(D)y=0$. Since $D^ke^{\lambda x}=\lambda^ke^{\lambda x}$,
$$
p(D)e^{\lambda x}=p(\lambda)e^{\lambda x}.
$$
The exponential never vanishes, so \b[$\boxed{e^{\lambda x}\text{ solves the equation }\Longleftrightarrow p(\lambda)=0}$].
For the second claim, the <product rule> gives
$$
D^k(xe^{\mu x})=e^{\mu x}(x\mu^k+k\mu^{k-1})
$$
for $k\geq1$, with the $k=0$ term treated separately. Summing the coefficients yields
$$
p(D)(xe^{\mu x})=e^{\mu x}\bigl(xp(\mu)+p'(\mu)\bigr).
$$
A root of multiplicity at least two has $p(\mu)=p'(\mu)=0$, proving \b[$xe^{\mu x}$ is also a solution]. This directly establishes the repeated-root contribution without assuming a solution formula. If the displayed coefficient list has leading zeros, the exponential calculation still applies to its actual <polynomial> degree.
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