# Find the Laplace transforms of the following {L}{leftlbracefrac{{{e}^{{-{6}{t}}} sinh{{left({2}{t}right)}}}}{{{e}^{{-{6}{t}}} cosh{{left({2}{t}right)}}}}rightrbrace}

Question
Laplace transform
Find the Laplace transforms of the following
$${L}{\left\lbrace\frac{{{e}^{{-{6}{t}}} \sinh{{\left({2}{t}\right)}}}}{{{e}^{{-{6}{t}}} \cosh{{\left({2}{t}\right)}}}}\right\rbrace}$$

2020-10-28
Step 1
We have to evaluate the Laplace transformation $${L}{\left\lbrace\frac{{{e}^{{-{6}{t}}} \sinh{{\left({2}{t}\right)}}}}{{{e}^{{-{6}{t}}} \cosh{{\left({2}{t}\right)}}}}\right\rbrace}$$
Use the property of Laplace transformation
$${L}{\left\lbrace\frac{{{e}^{{-{6}{t}}} \sinh{{\left({2}{t}\right)}}}}{{{e}^{{-{6}{t}}} \cosh{{\left({2}{t}\right)}}}}\right\rbrace}=\frac{{{L}{\left({e}^{{-{6}{t}}} \sinh{{\left({2}{t}\right)}}\right)}}}{{{L}{\left({e}^{{-{6}{t}}} \cosh{{\left({2}{t}\right)}}\right)}}}$$
Use the property
$${L}{\left({e}^{{{a}{t}}} \sinh{{\left({b}{t}\right)}}\right)}=\frac{b}{{{\left({s}-{a}\right)}^{2}-{b}^{2}}}\ \text{ and }\ {L}{\left({e}^{{{a}{t}}} \cosh{{\left({b}{t}\right)}}\right)}=\frac{{{s}-{a}}}{{{\left({s}-{a}\right)}^{2}-{b}^{2}}}$$
Step 2
Hence,
$${L}{\left({e}^{{-{6}{t}}} \sinh{{\left({2}{t}\right)}}\right)}=\frac{2}{{{\left({s}-{\left(-{6}\right)}\right)}^{2}-{2}^{2}}}$$
$$=\frac{2}{{{\left({s}+{6}\right)}^{2}-{4}}}$$
Hence, $${L}{\left({e}^{{-{6}{t}}} \sinh{{\left({2}{t}\right)}}\right)}=\frac{2}{{{\left({s}+{6}\right)}^{2}-{4}}}$$
And,
$${L}{\left({e}^{{-{6}{t}}} \cosh{{\left({2}{t}\right)}}\right)}=\frac{{{s}-{\left(-{6}\right)}}}{{{\left({s}-{\left(-{6}\right)}\right)}^{2}-{2}^{2}}}$$
$$=\frac{{{s}+{6}}}{{{\left({s}+{6}\right)}^{2}-{4}}}$$
Hence, $${L}{\left({e}{6}{\left(-{6}{t}\right)} \cosh{{\left({2}{t}\right)}}\right)}=\frac{{{s}+{6}}}{{{\left({s}+{6}\right)}^{2}-{4}}}$$
Step 3
Hence,$${L}{\left\lbrace\frac{{{e}^{{-{6}{t}}} \sinh{{\left({2}{t}\right)}}}}{{{e}^{{-{6}{t}}} \cosh{{\left({2}{t}\right)}}}}\right\rbrace}=\frac{{{\left(\frac{2}{{\left({s}+{6}\right)}^{2}}-{4}\right)}}}{{\frac{{{s}+{6}}}{{\left({s}+{6}\right)}^{2}}-{4}}}$$
$$=\frac{2}{{{s}+{6}}}$$
Hence, $${L}{\left\lbrace\frac{{{e}^{{-{6}{t}}} \sinh{{\left({2}{t}\right)}}}}{{{e}^{{-{6}{t}}} \cosh{{\left({2}{t}\right)}}}}\right\rbrace}=\frac{2}{{{s}+{6}}}$$

### Relevant Questions

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.
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) - ?
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) $$\displaystyle{L}{\left[{e}^{{-{3}{t}}}\right]}$$
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.)
$${L}{\left\lbrace{e}^{{-{t}}}\cdot{e}^{t} \cos{{\left({t}\right)}}\right\rbrace}$$
Find the Laplace transformation (evaluating the improper integral that defines this transformation) of the real valued function f(t) of the real variable t>0. (Assume the parameter s appearing in the Laplace transformation, as a real variable).
$$f{{\left({t}\right)}}={2}{t}^{2}-{4} \cosh{{\left({3}{t}\right)}}+{e}^{{{t}^{2}}}$$
Find the Laplace transforms of the given functions.
$$f{{\left({t}\right)}}={6}{e}^{{-{5}{t}}}+{e}^{{{3}{t}}}+{5}{t}^{{{3}}}-{9}$$
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 Laplace transform to find the solution of the IVP
$$2y'+y=0 , y(0)=-3$$
a) $$f{{\left({t}\right)}}={3}{e}^{{-{2}{t}}}$$
b)$$f{{\left({t}\right)}}={3}{e}^{{\frac{t}{{2}}}}$$
c)$$f{{\left({t}\right)}}={6}{e}^{{{2}{t}}} d) \(f{{\left({t}\right)}}={3}{e}^{{-\frac{t}{{2}}}}$$
$${y}\text{}-{4}{y}={e}^{{-{3}{t}}},{y}{\left({0}\right)}={0},{y}'{\left({0}\right)}={2}$$
a) $$\frac{14}{{20}}{e}^{{{2}{t}}}-\frac{5}{{30}}{e}^{{-{2}{t}}}-\frac{9}{{30}}{e}^{{-{6}{t}}}$$
b) $$\frac{11}{{20}}{e}^{{{2}{t}}}-\frac{51}{{20}}{e}^{{-{2}{t}}}-\frac{4}{{20}}{e}^{{-{3}{t}}}$$
c) $$\frac{14}{{15}}{e}^{{{2}{t}}}-\frac{5}{{10}}{e}^{{-{2}{t}}}-\frac{9}{{20}}{e}^{{-{3}{t}}}$$
d) $$\frac{14}{{20}}{e}^{{{2}{t}}}+\frac{5}{{20}}{e}^{{-{2}{t}}}-\frac{9}{{20}}{e}^{{-{3}{t}}}$$
$$f(t)=tu_2(t)$$
Ans. $$F(s)=\left(\frac{1}{s^2}+\frac{2}{s}\right)e^{-2s}$$