Question

In problems 1-20 find either F(s) or f(t), as indicated.

15.${\mathit{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{s}{s^2+4s+5}\right\}$

Short answer

The solution of the given Inverse Laplace transform is$e^{-2t}\cos \left(t\right)-2e^{-2t}\sin \left(t\right)$.

Step-by-step solution

Definition of Inverse transforms

If $\mathit{F}\mathbf{\left(}\mathit{s}\mathbf{\right)}$represents the Laplace transform of a function$\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}$, that is, $\mathit{L}\left\{f\left(t\right)\right\}\mathbf{=}\mathit{F}\mathbf{\left(}\mathit{s}\mathbf{\right)}$,we then say$\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}$is the inverse Laplace transform of$\mathit{F}\mathbf{\left(}\mathit{s}\mathbf{\right)}$and write $\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathit{L}}^{\mathbf{-}\mathbf{1}}\left\{F\left(s\right)\right\}$.

Inverse Laplace transform of given function

$L^{-1}\left\{\frac{s}{s^2+4s+5}\right\}$

Complete the square within the denominator to get this Inverse Transformation into a usable form.

$L^{-1}\left\{\frac{s}{s^2+4s+5}\right\}=L^{-1}\left\{\frac{s}{\left(s+2\right)\left(s+2\right)+1}\right\}$

We presume it is the Inverse of a trigonometric function (specifically $\left(\cos t\right)$) because of the in the denominator and the in the numerator as below.

$\Rightarrow L^{-1}\left\{\frac{s}{{\left(s+2\right)}^2+1}\right\}$

To compute the Inverse of $L^{-1}\left\{\frac{s}{{\left(s+2\right)}^2+1}\right\}$, we need the term in the numerator to

equal theterm in the denominator.

By adding and subtracting 2, $L^{-1}\left\{\frac{s}{{\left(s+2\right)}^2+1}\right\}=L^{-1}\left\{\frac{s+2-2}{{\left(s+2\right)}^2+1}\right\}$

Step 2: Compute the Inverse Transformation

Expand the numerator into a usable form,

$\begin{array}{rcl}& \Rightarrow & L^{-1}\left\{\frac{s+2-2}{{\left(s+2\right)}^2+1}\right\}=L^{-1}\left\{\frac{s+2}{{\left(s+2\right)}^2+1}-\frac{2}{{\left(s+2\right)}^2+1}\right\}\\ & =& L^{-1}\left\{\frac{s+2}{{\left(s+2\right)}^2+1}\right\}-2L^{-1}\left\{\frac{1}{{\left(s+2\right)}^2+1}\right\}\end{array}$

Now, Compute the Inverse Transformation,

$=e^{-2t}\cos \left(t\right)-2e^{-2t}\sin \left(t\right)$

Thus, $L^{-1}\left\{\frac{s}{s^2+4s+5}\right\}=e^{-2t}\cos \left(t\right)-2e^{-2t}\sin \left(t\right)$

Therefore, the solution of the given Inverse Laplace transform $L^{-1}\left\{\frac{s}{s^2+4s+5}\right\}$is $e^{-2t}\cos \left(t\right)-2e^{-2t}\sin \left(t\right)$.