Calculate the inverse Laplace transform $\frac{s-2}{(s+1)^4}$

Question

I need to calculate the inverse Laplace transform of $$\frac{s-2}{(s+1)^4}$$

Not quite sure how to do this one. I see that you should break the numerator up into $$\frac{s}{(s+1)^4}-\frac{2}{(s+1)^4}$$.

But looking at my table of elementary inverse inverse laplace transforms I can't quite figure out how to manipulate the bottom to get a correct answer

Answer 1

First, we should find the inverse Laplace transform of $\frac{1}{(s+1)^4}$. If you look at a table of Laplace transforms, you'll see that $$L(t^ne^{at}) = \frac{n!}{(s-a)^{n+1}}$$ (this formula can be shown by induction & integration by parts). So we can see that $$L(\frac{t^3e^{-t}}{3!}) = \frac{1}{(s+1)^4}$$

Most tables will also mention that $$L(f'(t))(s) = s\cdot L(f(t)) - f(0)$$ And so we have $$L(3t^2e^{-t} - t^3e^{-t}) = L(\frac{d}{dt}t^3e^{-t}) = \frac{3!s}{(s+1)^4}$$ And so the inverse Laplace transform of our original function is

$$\frac{1}{3!}(3t^2e^{-t} - t^3e^{-t} - 2t^3e^{-t})$$

Answer 2

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Note that $\ds{{s - 2 \over \pars{s + 1}^{4}} = {1 \over \pars{s + 1}^{3}} - {3 \over \pars{s + 1}^{4}}}$.

With $\ds{\ul{\sigma, \mu > 0}}$: \begin{align} \,\mathrm{f}\pars{\mu} & \equiv \int_{\sigma - \infty\ic}^{\sigma + \infty\ic}{\expo{st} \over s + \mu}\, {\dd s \over 2\pi\ic} = \expo{-\mu t} \end{align}

---

$$ \left\lbrace\begin{array}{rcccl} \ds{\int_{\sigma - \infty\ic}^{\sigma + \infty\ic} {\expo{st} \over \pars{s + 1}^{3}}\,{\dd s \over 2\pi\ic}} & \ds{=} & \ds{\half\,\,\mathrm{f}''\pars{1}} & \ds{=} & \ds{\half\,t^{2}\expo{-t}} \\[3mm] \ds{\int_{\sigma - \infty\ic}^{\sigma + \infty\ic} {\expo{st} \over \pars{s + 1}^{4}}\,{\dd s \over 2\pi\ic}} & \ds{=} & \ds{-\,{1 \over 6}\,\,\mathrm{f}'''\pars{1}} & \ds{=} & \ds{{1 \over 6}\,t^{3}\expo{-t}} \end{array}\right. $$

---

$$ \color{#f00}{Solution}:\quad \half\,t^{2}\expo{-t} - 3\pars{{1 \over 6}\,t^{3}\expo{-t}} = \color{#f00}{\half\,t^{2}\pars{1 - t}\expo{-t}} $$