Inverse Laplace transform of $\frac{s^2}{s^2+4s+13}$

Question

Would you know where I was wrong?

$$\mathcal{L}^{-1}\left[ \dfrac{s^2}{s^2+4s+13}\right] =\mathcal{L}^{-1}\left[ \dfrac{s^2}{(s+2)^2+9}\right] =-\dfrac{d}{ds}\mathcal{L}^{-1}\left[ \dfrac{s}{(s+2)^2+9}\right] =\dfrac{-de^{-2t}}{ds}\mathcal{L}^{-1}\left[ \dfrac{s-2}{s^2+9}\right]\left. \right|_{s\to s+2}=-\dfrac{de^{-2t}}{ds}\left( \mathcal{L}^{-1}\left[ \dfrac{s}{s^2+9}\right] -2\mathcal{L}^{-1}\left[ \dfrac{1}{s^2+9}\right] \right) =\dfrac{-d}{ds}e^{-2t}\left( \cos (3t)-\dfrac23 \sin (3t)\right) =4e^{-2t}\cos (3t)+\dfrac53 e^{-2t}\sin (3t)$$ The solution they give me is: $$\boxed{g(t)=-4e^{-2t}\cos (3t)-\dfrac53 e^{-2t}\sin (3t)+\delta (t).}$$

Answer 1

Hint.

$$ \dfrac{s^2}{s^2+4s+13} = 1-\dfrac{4s+13}{s^2+4s+13} $$

NOTE

$$ \dfrac{s^2}{s^2+4s+13} \ \ \ \text{is not strictly causal} $$