Question

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

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

Short answer

The solution of the given Inverse Laplace transform is $\frac{1}{2}{e}^{-t}\sin 2t$

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 $\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$is the inverse Laplace transform of $\mathit{F}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$and write localid="1664219065168" $\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathit{L}}^{\mathbf{-}\mathbf{1}}\mathbf{\left\{}\mathbf{F}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\right\}}$.

Inverse Laplace transform of given function

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

Use first shifting theorem

By using Inverse form of the first shifting theorem

$L^{-1}\left\{F(s-a)\right\}=L^{-1}\left\{{\overline{)F\left(s\right)}}_{s\to s-a}\right\}=e^{at}f\left(t\right)$

Completion of square,

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

$=\frac{1}{2}{L}^{-1}\left\{{\overline{)\frac{2}{s^2+2^2}}}_{s\to s-(-1)}\right\}$ [using inverse form of first shifting theorem]

$=\frac{1}{2}{e}^{-t}\sin 2t$

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