# Find the inverse Laplace transform f{{left({t}right)}}={L}^{ -{{1}}}{leftlbrace{F}{left({s}right)}rightrbrace} 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.

Question
Laplace transform
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.

2020-11-02
$${\left({i}\right)}{F}{\left({s}\right)}=\frac{{{2}{s}+{1}}}{{{s}^{2}-{2}{s}+{1}}}$$
$$\frac{{{2}{s}+{1}}}{{\left({s}-{1}\right)}^{2}}=\frac{A}{{{s}-{1}}}+\frac{B}{{\left({s}-{1}\right)}^{2}}$$
$${2}{s}+{1}=\frac{A}{{{s}-{1}}}+\frac{B}{{\left({s}-{1}\right)}^{2}}$$
$$2s+1=A(s-1)+B$$
$$A=2, -A+B=1$$
$$B=3$$
$${F}{\left({s}\right)}=\frac{2}{{{s}-{1}}}+\frac{3}{{\left({s}-{1}\right)}^{2}}$$
$$f{{\left({t}\right)}}={2}{e}^{t}+{3}{t}{e}^{t} \({\left({i}{i}\right)}{F}{\left({s}\right)}=\frac{{{3}{s}+{2}}}{{{s}^{2}-{3}{s}+{2}}}$$
$$\frac{{{3}{s}+{2}}}{{{\left({s}-{2}\right)}{\left({s}-{1}\right)}}}=\frac{A}{{{s}-{2}}}+\frac{B}{{{s}-{1}}}$$
$$3s+2=A(s-1)+B(s-2)$$
$$A+B=3$$
$$A=8 , B=-5$$
$$-A-2B=2$$
$${F}{\left({s}\right)}=\frac{8}{{{s}-{2}}}-\frac{5}{{{s}-{1}}}$$
$$f{{\left({t}\right)}}={8}{e}^{{{2}{t}}}-{5}{e}^{{{t}}}$$
$${\left({i}{i}{i}\right)}{F}{\left({s}\right)}=\frac{{{3}{s}^{2}+{4}}}{{{\left({s}^{2}+{1}\right)}{\left({s}-{1}\right)}}}$$
$$\frac{{{3}{s}^{2}+{4}}}{{{\left({s}-{1}\right)}{\left({s}^{2}+{1}\right)}}}=\frac{{{A}{s}+{B}}}{{{s}^{2}+{1}}}+\frac{C}{{{s}-{1}}}$$
$$\Rightarrow{3}{s}^{2}+{4}={A}{\left({s}^{2}-{s}\right)}+{B}{\left({s}-{1}\right)}+{C}{\left({s}^{2}+{1}\right)}$$
$$A+B=3$$
$$-A+B=0$$
$$-B+C=4$$
$${A}=-\frac{1}{{2}},{B}=-\frac{1}{{2}},{C}=\frac{7}{{2}}$$
$${F}{\left({s}\right)}=\frac{{-{1}{s}}}{{{2}{\left({s}^{2}+{1}\right)}}}-\frac{1}{{{2}{\left({s}^{2}+{1}\right)}}}+\frac{7}{{{2}{\left({s}-{1}\right)}}}$$
$$f{{\left({t}\right)}}=-\frac{1}{{2}} \cos{{t}}-\frac{1}{{2}} \sin{{t}}+\frac{7}{{2}}{e}^{t}$$

### Relevant Questions

In an integro-differential equation, the unknown dependent variable y appears within an integral, and its derivative $$\frac{dy}{dt}$$ also appears. Consider the following initial value problem, defined for t > 0:
$$\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}+{4}{\int_{{0}}^{{t}}}{y}{\left({t}-{w}\right)}{e}^{{-{4}{w}}}{d}{w}={3},{y}{\left({0}\right)}={0}$$
a) Use convolution and Laplace transforms to find the Laplace transform of the solution.
$${Y}{\left({s}\right)}={L}{\left\lbrace{y}{\left({t}\right)}\right)}{\rbrace}-?$$
b) Obtain the solution y(t).
y(t) - ?
Use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the convolution integral before transforming.(Write your answer as a function of s.)
$${L}{\left\lbrace{e}^{{-{t}}}\cdot{e}^{t} \cos{{\left({t}\right)}}\right\rbrace}$$
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, $$L^{-1}[F(s)] = f(t)$$
Use the definition of Laplace Transforms to show that:
\displaystyle{L}{\left\lbrace{t}^{n}\right\rbrace}=\frac{{{n}!}}{{{s}^{{{n}+{1}}}}},{n}={1},{2},{3},\ldots\)
Use the appropriate algebra and Table of Laplace's Transform to find the given inverse Laplace transform. $$L^{-1}\left\{\frac{1}{(s-1)^2}-\frac{120}{(s+3)^6}\right\}$$
Find the Laplace transforms of the following time functions.
Solve problem 1(a) and 1 (b) using the Laplace transform definition i.e. integration. For problem 1(c) and 1(d) you can use the Laplace Transform Tables.
a)$$f(t)=1+2t$$ b)$$f(t) =\sin \omega t \text{Hint: Use Euler’s relationship, } \sin\omega t = \frac{e^(j\omega t)-e^(-j\omega t)}{2j}$$
c)$$f(t)=\sin(2t)+2\cos(2t)+e^{-t}\sin(2t)$$
Find the laplace transform by definition.
a) $$\displaystyle{L}{\left\lbrace{2}\right\rbrace}$$
b) $$\displaystyle{L}{\left\lbrace{e}^{{{2}{t}}}\right\rbrace}$$
c) $$\displaystyle{L}{\left[{e}^{{-{3}{t}}}\right]}$$
Please solve the 2nd order differential equation by (PLEASE FOLLOW GIVEN METHOD) LAPLACE TRANSFORMATION
ALSO, USE PARTIAL FRACTION WHEN YOU ARRIVE
$$L(y) = \left[\frac{w}{(s^2 + a^2)(s^2+w^2)}\right]*b$$
Problem 2 Solve the differential equation
$$\frac{d^2y}{dt^2}+a^2y=b \sin(\omega t)$$ where $$y(0)=0$$
and $$y'(0)=0$$
The diagrams are: $$(1) {1,i,-i}. (2) {-1+4i,-1-4i}. (3) {-1}. (4)$$ The empty diagram.
(b) A mechanical system is discovered during an archaeological dig in Ethiopia. Rather than break it open, the investigators subjected it to a unit impulse. It was found that the motion of the system in response to the unit impulse is given by $$w(t) = u(t)e^{-\frac{t}{2}} \sin(\frac{3t}{2})$$
$${L}{\left\lbrace\frac{{{e}^{{-{6}{t}}} \sinh{{\left({2}{t}\right)}}}}{{{e}^{{-{6}{t}}} \cosh{{\left({2}{t}\right)}}}}\right\rbrace}$$