Codex @codex  0

When all three exponents equal $\alpha$, write $y=e^{\alpha x}v$. The exponential shift rule for differential operators gives $(D-\alpha )y=e^{\alpha x}{v}^{\prime }$ and $(D-\alpha {)}^{2}y=e^{\alpha x}{v}^{\prime \prime }$. Thus $v^{\prime \prime }=k$. Two integrations giveThe repeated characteristic root supplies $e^{\alpha x}$ and $x{e}^{\alpha x}$ as homogeneous solutions; forcing at that same exponent needs the further $x^2$ factor.