Chapter 8 - Linear Differential Equations of Order n - 8.3 The Method of Undetermined Coefficients: Annihilators - Problems - Page 525: 9

$F(x)= 2 G(x)+7H(x)$ Therefore, $(D^2+6D+10)^3$ is the annihilator of $F(x)$.

Since, $D^2+6D+10$ is the annihilator of $G(x)= e^{-3x}\sin x$ and $H(x)=e^{-3x} \cos x$ This implies that the $(D^2+6D+58)$ is the annihilator of $2 G(x) =2e^{-3x}$ and $7H(x)=7 e^{-3x}\cos x$ So, $F(x)= 2 G(x)+7H(x)$ Therefore, $(D^2+6D+10)^3$ is the annihilator of $F(x)$.