Question

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

19.${\mathit{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{2s-1}{s^2{\left(s+1\right)}^3}\right\}$

Short answer

The solution of the given Inverse Laplace transform is $5-t-5e^{-t}-4t{e}^{-t}-\frac{3}{2}{t}^2{e}^{-t}$.

Step-by-step solution

Definition of Inverse transforms

If $\mathit{F}\mathbf{\left(}\mathit{s}\mathbf{\right)}$represents the Laplace transform of a function role="math" localid="1664225095882" $\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}$, 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

$L^{-1}\left\{\frac{2s-1}{s^2{\left(s+1\right)}^3}\right\}$

By using partial function,

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

Multiply by$s^2{\left(s+1\right)}^3$ on both sides,

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

$=s^4\left(A+C\right)+s^3\left(3A+B+2C+D\right)+s^2\left(3A+3B+C+D+E\right)+S\left(A+3B\right)+B$

After comparing respective order of coefficient,

$\begin{array}{l}B=-1\\ A+3B=2\\ A+C=0\\ 3A+3B+2C+D=0\\ 3A+3B+C+D+E=0\end{array}$ $\begin{array}{l}\left(B=-1\right)\\ \left(A=5\right)\\ \left(C=-5\right)\\ \left(D=-4\right)\\ \left(E=-3\right)\end{array}$

Inverse form of first shifting theorem

By using Inverse form of the first shifting theorem

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

$\frac{2s-1}{s^2{\left(s+1\right)}^3}=\frac{5}{s}-\frac{1}{{\left(s\right)}^2}-\frac{5}{\left(s+1\right)}-\frac{4}{{\left(s+1\right)}^2}-\frac{3}{{\left(s+1\right)}^3}$

Take $L^{-1}$on both sides,

$\begin{array}{l}{L}^{-1}\left\{\frac{2s-1}{s^2{\left(s+1\right)}^3}\right\}=5L^{-1}\left\{\frac{1}{s}\right\}-L^{-1}\left\{\frac{1}{{\left(s\right)}^2}\right\}-5L^{-1}\left\{\frac{1}{\left(s+1\right)}\right\}-4L^{-1}\left\{\frac{1}{{\left(s+1\right)}^2}\right\}-3L^{-1}\left\{\frac{1}{{\left(s+1\right)}^3}\right\}\\ \end{array}$

Use Inverse form of the first shifting theorem,

$=5-t-5e^{-t}-4t{e}^{-t}-\frac{3}{2}{t}^2{e}^{-t}$

In the process above, we used

$$\begin{gathered} \begin{array}{l}{L}^{-1}\left\{\frac{1}{s}\right\}=1\\ L^{-1}\left\{\frac{1}{{\left(s\right)}^2}\right\}=t\\ L^{-1}\left\{\frac{1}{\left(s+1\right)}\right\}=L^{-1}\left\{{\overline{)\frac{1}{s}}}_{s\to s+1}\right\}=e^{-t} \\ \\ L^{-1}\left\{\frac{1}{{\left(s+1\right)}^2}\right\}==L^{-1}\left\{{\overline{)\frac{1}{s^2}}}_{s\to s+1}\right\}=t{e}^{-t}\\ L^{-1}\left\{\frac{1}{{\left(s+1\right)}^3}\right\}=L^{-1}\left\{{\overline{)\frac{{\displaystyle 1}}{{\displaystyle s^3}}}}_{s\to s+1}\right\}=t^2{e}^{-t}\end{array} \end{gathered}$$

Thus, $L^{-1}\left\{\frac{2s-1}{s^2{\left(s+1\right)}^3}\right\}=5-t-5e^{-t}-4t{e}^{-t}-\frac{3}{2}{t}^2{e}^{-t}$

Therefore, the solution of the given Inverse Laplace transform $L^{-1}\left\{\frac{2s-1}{s^2{\left(s+1\right)}^3}\right\}$is data-custom-editor="chemistry" $5-t-5e^{-t}-4t{e}^{-t}-\frac{3}{2}{t}^2{e}^{-t}$.