Question

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

18.${\mathit{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{5s}{{\left(s-2\right)}^2}\right\}$

Short answer

The solution of the given Inverse Laplace transform is $5e^{2t}\left(1+2t\right)$.

Step-by-step solution

Definition of Inverse transforms

$\mathit{L}\left\{f\left(t\right)\right\}\mathbf{=}\mathit{F}\mathbf{\left(}\mathit{s}\mathbf{\right)}$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, ,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

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

Inverse form of first shifting theorem

By using Inverse form of the first shifting theorem

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

Adding zero in the numerator,

$L^{-1}\left\{\frac{5s}{{\left(s-2\right)}^2}\right\}=L^{-1}\left\{\frac{5s-10+10}{{\left(s-2\right)}^2}\right\}$

$=5L^{-1}\left\{\frac{s-2}{{\left(s-2\right)}^2}\right\}+L^{-1}\left\{\frac{10}{{\left(s-2\right)}^2}\right\}$ [using term-wise division]

$=5L^{-1}\left\{{\overline{)\frac{1}{s}}}_{s\to s-2}\right\}+10L^{-1}\left\{{\overline{)\frac{{\displaystyle 1}}{{\displaystyle s^2}}}}_{s\to s-2}\right\}$

Use Inverse form of the first shifting theorem,

$$\begin{gathered} =5e^{2t}+10t{e}^{2t} \\ =5e^{2t}\left(1+2t\right) \end{gathered}$$

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

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