Problem 03 | Inverse Laplace Transform

$\mathcal{L}^{-1} \left[ \dfrac{7}{s^2 + 6} \right] = 7 \mathcal{L}^{-1} \left[ \dfrac{1}{s^2 + (\sqrt{6})^2} \right]$

$\mathcal{L}^{-1} \left[ \dfrac{7}{s^2 + 6} \right] = 7 \times \dfrac{1}{\sqrt{6}}\sin \sqrt{6}t$

$\mathcal{L}^{-1} \left[ \dfrac{7}{s^2 + 6} \right] = \dfrac{7}{\sqrt{6}} \times \dfrac{\sqrt{6}}{\sqrt{6}} \sin \sqrt{6}t$

$\mathcal{L}^{-1} \left[ \dfrac{7}{s^2 + 6} \right] = \dfrac{7\sqrt{6}}{6}\sin \sqrt{6}t$           answer