# How should I inverse Laplace transform this function? ((omega_n)^2)/(s^2+bs+c)

How should I inverse Laplace transform this function?
$\frac{\left({\omega }_{n}{\right)}^{2}}{{s}^{2}+bs+c}$
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In such a case the most convenient way to compute the inverse transform is to use standard results from Laplace transform tables. The transform pairs that you need here can be found in most tables:
$\frac{1}{{s}^{2}}⟺t\cdot u\left(t\right)\phantom{\rule{0ex}{0ex}}\frac{{\omega }_{0}}{{s}^{2}+{\omega }_{0}^{2}}⟺\mathrm{sin}\left({\omega }_{0}t\right)\cdot u\left(t\right)\phantom{\rule{0ex}{0ex}}\frac{{\omega }_{0}}{{s}^{2}-{\omega }_{0}^{2}}⟺\mathrm{sinh}\left({\omega }_{0}t\right)\cdot u\left(t\right)\phantom{\rule{0ex}{0ex}}f\left(t\right)⟺F\left(s\right)\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}{e}^{-\alpha t}f\left(t\right)⟺F\left(s+\alpha \right)$
Now rewrite the function as
$X\left(s\right)=\frac{{\omega }_{n}^{2}}{{s}^{2}+bs+c}=\frac{{\omega }_{n}^{2}}{\left(s+b/2{\right)}^{2}+c-{b}^{2}/4}$
You have to distinguish 3 cases:
$c>{b}^{2}/4$: Define ${\omega }_{0}^{2}=c-{b}^{2}/4$
$X\left(s\right)=\frac{{\omega }_{n}^{2}}{{\omega }_{0}}\frac{{\omega }_{0}}{\left(s+b/2{\right)}^{2}+{\omega }_{0}^{2}}⟺\frac{{\omega }_{n}^{2}}{{\omega }_{0}}{e}^{-bt/2}\mathrm{sin}\left({\omega }_{0}t\right)\cdot u\left(t\right)$
$c<{b}^{2}/4$: Define ${\omega }_{0}^{2}={b}^{2}/4-c$
$X\left(s\right)=\frac{{\omega }_{n}^{2}}{{\omega }_{0}}\frac{{\omega }_{0}}{\left(s+b/2{\right)}^{2}-{\omega }_{0}^{2}}⟺\frac{{\omega }_{n}^{2}}{{\omega }_{0}}{e}^{-bt/2}\mathrm{sinh}\left({\omega }_{0}t\right)\cdot u\left(t\right)$
$c={b}^{2}/4$
$X\left(s\right)={\omega }_{n}^{2}\frac{1}{\left(s+b/2{\right)}^{2}}⟺{\omega }_{n}^{2}{e}^{-bt/2}t\cdot u\left(t\right)$