Question

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

20.${\mathit{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{{\left(s+1\right)}^2}{{\left(s+2\right)}^4}\right\}$

Short answer

The solution of the given Inverse Laplace transform is $e^{-2t}\left(t-2t^2+\frac{1}{6}{t}^3\right)$.

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

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

By using partial function,

$L^{-1}\left\{\frac{{\left(s+1\right)}^2}{{\left(s+2\right)}^4}\right\}=\frac{A}{s+2}+\frac{B}{{\left(s+2\right)}^2}+\frac{C}{{\left(s+2\right)}^3}+\frac{D}{{\left(s+2\right)}^4}$

Evaluate unknown coefficient A,B,C and D.

First, evaluate D by multiply${\left(s+2\right)}^4$ in both sides and put$s=-2$

${\left(s+1\right)}^2\left|{}_{s=-2}=D\right.,D=1$ ,

${\left(s+1\right)}^2-1=A{\left(s+2\right)}^3+B{\left(s+2\right)}^2+C\left(s+2\right)$ [after simplify]

$s^2-2s+1=A{s}^3+\left(3A+B\right)s^2+\left(3A+2B+C\right)s+A+B+C$

After comparing respective order of coefficient,

$A=0,3A+B=1,B=1,3A+2B+C=-2,C=-4$

Inverse form of first shifting theorem

By using Inverse form of the first shifting theorem

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

$\begin{array}{rcl}{L}^{-1}\left\{\frac{{\left(s+1\right)}^2}{{\left(s+2\right)}^4}\right\}& =& L^{-1}\left\{\frac{1}{{\left(s+2\right)}^2}-\frac{4}{{\left(s+2\right)}^3}+\frac{1}{{\left(s+2\right)}^4}\right\}\\ & =& L^{-1}\left\{{\overline{)\frac{1}{s}}}_{s\to s-(-2)}\right\}+2L^{-1}\left\{{\overline{)\frac{{\displaystyle 2!}}{{\displaystyle s^2}}}}_{s\to s-(-2)}\right\}+\frac{1}{3!}{L}^{-1}\left\{{\overline{)\frac{{\displaystyle 3!}}{{\displaystyle s^3}}}}_{s\to s-(-2)}\right\}\end{array}$

Use Inverse form of the first shifting theorem,

$$\begin{gathered} =t{e}^{-2t}-2t^2{e}^{-2t}+\frac{1}{6}{t}^3{e}^{-2t} \\ =e^{-2t}\left(t-2t^2+\frac{1}{6}{t}^3\right) \end{gathered}$$

Thus, $L^{-1}\left\{\frac{{\left(s+1\right)}^2}{{\left(s+2\right)}^4}\right\}=e^{-2t}\left(t-2t^2+\frac{1}{6}{t}^3\right)$

Therefore, the solution of the given Inverse Laplace transform $L^{-1}\left\{\frac{{\left(s+1\right)}^2}{{\left(s+2\right)}^4}\right\}$ is $e^{-2t}\left(t-2t^2+\frac{1}{6}{t}^3\right)$.