# Find the inverse Laplace transform displaystylefrac{{{4}{s}^{3}-{s}^{2}+{5}{s}+{2}}}{{{s}^{4}}}

Question
Laplace transform
Find the inverse Laplace transform
$$\displaystyle\frac{{{4}{s}^{3}-{s}^{2}+{5}{s}+{2}}}{{{s}^{4}}}$$

2021-02-09
Step 1
If $$\displaystyle{A}{\left({s}\right)}={L}{\left\lbrace{B}{\left({t}\right)}\right\rbrace}$$, then the inverse transform of A(s) is defined as:
$$\displaystyle{L}^{{-{1}}}{\left[{A}{\left({s}\right)}\right]}={B}{\left({t}\right)}$$
Laplace transform of a function f(t) is given by:
$$\displaystyle{L}{\left\lbrace f{{\left({t}\right)}}\right\rbrace}={\int_{{0}}^{\infty}}{e}^{{-{s}{t}}} f{{\left({t}\right)}}{\left.{d}{t}\right.}$$
Step 2
$$\displaystyle{L}^{ -{{1}}}{\left(\frac{{{4}{s}^{3}-{s}^{2}+{5}{s}+{2}}}{{{s}^{4}}}\right)}$$
$$\displaystyle={L}^{ -{{1}}}{\left(\frac{4}{{s}}-\frac{1}{{s}^{2}}+\frac{5}{{s}^{3}}+\frac{2}{{s}^{4}}\right)}$$
$$\displaystyle={L}^{ -{{1}}}{\left(\frac{4}{{s}}\right)}-{L}^{ -{{1}}}{\left(\frac{1}{{s}^{2}}\right)}+{L}^{ -{{1}}}{\left(\frac{5}{{s}^{3}}\right)}+{L}^{ -{{1}}}{\left(\frac{2}{{s}^{4}}\right)}$$
$$\displaystyle={4}{H}{\left({t}\right)}-{t}+\frac{{{5}{t}^{2}}}{{2}}+\frac{{{t}^{3}}}{{3}}$$

### Relevant Questions

Write down the qualitative form of the inverse Laplace transform of the following function. For each question first write down the poles of the function , X(s)
a) $$X(s)=\frac{s+1}{(s+2)(s^2+2s+2)(s^2+4)}$$
b) $$X(s)=\frac{1}{(2s^2+8s+20)(s^2+2s+2)(s+8)}$$
c) $$X(s)=\frac{1}{s^2(s^2+2s+5)(s+3)}$$
Use properties of the Laplace transform to answer the following
(a) If $$f(t)=(t+5)^2+t^2e^{5t}$$, find the Laplace transform,$$L[f(t)] = F(s)$$.
(b) If $$f(t) = 2e^{-t}\cos(3t+\frac{\pi}{4})$$, find the Laplace transform, $$L[f(t)] = F(s)$$. HINT:
$$\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha) \sin(\beta)$$
(c) If $$F(s) = \frac{7s^2-37s+64}{s(s^2-8s+16)}$$ find the inverse Laplace transform, $$L^{-1}|F(s)| = f(t)$$
(d) If $$F(s) = e^{-7s}(\frac{1}{s}+\frac{s}{s^2+1})$$ , find the inverse Laplace transform, $$L^{-1}[F(s)] = f(t)$$
Find the inverse Laplace transform $$f{{\left({t}\right)}}={L}^{ -{{1}}}{\left\lbrace{F}{\left({s}\right)}\right\rbrace}$$ of each of the following functions.
$${\left({i}\right)}{F}{\left({s}\right)}=\frac{{{2}{s}+{1}}}{{{s}^{2}-{2}{s}+{1}}}$$
Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.
$${\left({i}{i}\right)}{F}{\left({s}\right)}=\frac{{{3}{s}+{2}}}{{{s}^{2}-{3}{s}+{2}}}$$
Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.
$${\left({i}{i}{i}\right)}{F}{\left({s}\right)}=\frac{{{3}{s}^{2}+{4}}}{{{\left({s}^{2}+{1}\right)}{\left({s}-{1}\right)}}}$$
Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.
find the inverse Laplace transform of the given function.
$${F}{\left({s}\right)}=\frac{{{2}{s}-{3}}}{{{s}^{2}-{4}}}$$
Find the inverse Laplace transform of $$F(s)=\frac{(s+4)}{(s^2+9)}$$
a)$$\cos(t)+\frac{4}{3}\sin(t)$$
b)non of the above
c) $$\cos(3t)+\sin(3t)$$
d) $$\cos(3t)+\frac{4}{3} \sin(3t)$$
e)$$\cos(3t)+\frac{2}{3} \sin(3t)$$
f) $$\cos(t)+4\sin(t)$$
find the inverse of Laplace transform
$$\frac{3}{(s+2)^2}-\frac{2s+6}{(s^2+4)}$$
Find the inverse of Laplace transform
$$\frac{2s+3}{(s-7)^4}$$
Find the inverse Laplace transform of the given function by using the convolution theorem. $${F}{\left({s}\right)}=\frac{s}{{{\left({s}+{1}\right)}{\left({s}^{2}+{4}\right)}}}$$
Find the Laplace transform $$L\left\{u_3(t)(t^2-5t+6)\right\}$$
$$a) F(s)=e^{-3s}\left(\frac{2}{s^4}-\frac{5}{s^3}+\frac{6}{s^2}\right)$$
$$b) F(s)=e^{-3s}\left(\frac{2}{s^3}-\frac{5}{s^2}+\frac{6}{s}\right)$$
$$c) F(s)=e^{-3s}\frac{2+s}{s^4}$$
$$d) F(s)=e^{-3s}\frac{2+s}{s^3}$$
$$e) F(s)=e^{-3s}\frac{2-11s+30s^2}{s^3}$$
Find inverse Laplace transform $$L^{-1}\left\{\frac{s-5}{s^2+5s-24}\right\}$$ Please provide supporting details for your answer