# To find: The Laplace transform of the function {L}{leftlbrace{t}^{4}-{t}^{2}-{t}+ sin{sqrt{{{2}{t}}}}rightrbrace}

Question
Transformation properties
To find: The Laplace transform of the function
$${L}{\left\lbrace{t}^{4}-{t}^{2}-{t}+ \sin{\sqrt{{{2}{t}}}}\right\rbrace}$$

2021-02-13
Approach:
With the help of Laplace Transform table and linearity property of Laplace Transformation calculation is carried out as follows:
Calculation:
$${L}{\left\lbrace{t}^{4}-{t}^{2}-{t}+ \sin{\sqrt{{{2}{t}}}}\right\rbrace}$$
Consider $$f,\ f_{1}\ \text{and}\ f_{2}\ \text{be the functions whose Laplace Transform exists for}\ s\ >\ \alpha$$ and c be a constant.
Then, for $$s\ >\ \alpha$$
$${L}{\left\lbrace{{f}_{{1}}+}{f}_{{2}}\right\rbrace}={L}{\left\lbrace{f}_{{1}}\right\rbrace}+{L}{\left\lbrace{f}_{{2}}\right\rbrace}$$
Also,
$${L}{\left\lbrace{c}{f}\right\rbrace}={c}{L}{\left\lbrace{f}\right\rbrace}$$
Using above properties:
$${L}{\left\lbrace{t}^{4}-{t}^{2}-{t}+ \sin{\sqrt{{{2}{t}}}}\right\rbrace}={L}{\left\lbrace{t}^{4}\right\rbrace}+{L}{\left\lbrace-{t}^{2}\right\rbrace}+{L}{\left\lbrace-{t}\right\rbrace}+{L}{\left\lbrace \sin{\sqrt{{{2}{t}}}}\right\rbrace}$$
$$={L}{\left\lbrace{t}^{4}\right\rbrace}-{L}{\left\lbrace{t}^{2}\right\rbrace}-{L}{\left\lbrace{t}\right\rbrace}+{L}{\left\lbrace \sin{\sqrt{{{2}{t}}}}\right\rbrace}$$
Since the Laplace Transform from the Laplace Transform table:
$${L}{\left\lbrace{t}^{n}\right\rbrace}=\frac{{{n}!}}{{s}^{{{n}+{1}}}},\forall{s}>{0},{n}={1},{2},{3},\ldots$$
$${L}{\left\lbrace \sin{{a}}{t}\right\rbrace}=\frac{a}{{{s}^{2}+{a}^{2}}},\forall{s}>{0}$$
Using above formulae:
$${L}{\left\lbrace{t}^{4}\right\rbrace}-{L}{\left\lbrace{t}^{2}\right\rbrace}-{L}{\left\lbrace{t}\right\rbrace}+{L}{\left\lbrace \sin{\sqrt{{{2}{t}}}}\right\rbrace}=\frac{{{4}!}}{{s}^{{{4}+{1}}}}-\frac{{{2}!}}{{s}^{{{2}+{1}}}}-\frac{{{1}!}}{{s}^{{{1}+{1}}}}+\frac{\sqrt{{2}}}{{{s}^{2}+{\left(\sqrt{{2}}\right)}^{2}}},\forall{s}>{0}$$
$$=\frac{24}{{s}^{5}}-\frac{2}{{s}^{3}}-\frac{1}{{s}^{2}}+\sqrt{/}{\left({s}^{2}+{2}\right)},\forall{s}>{0}$$
Gives, the Laplace Transform is
$${F}{\left({s}\right)}=\frac{24}{{s}^{5}}-\frac{2}{{s}^{3}}-\frac{1}{{s}^{2}}+\sqrt{/}{\left({s}^{2}+{2}\right)},\forall{s}>{0}$$
Conclusion:
Therefore, the required Laplace Transform for the given function is
$${F}{\left({s}\right)}=\frac{24}{{s}^{5}}-\frac{2}{{s}^{3}}-\frac{1}{{s}^{2}}+\sqrt{/}{\left({s}^{2}+{2}\right)},\forall{s}>{0}$$

### Relevant Questions

The Laplace transform of the function $${L}{\left\lbrace{6}{e}^{{-{3}{t}}}-{t}^{2}+{2}{t}-{8}\right\rbrace}$$
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 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 laplace transform by definition.
a) $$\displaystyle{L}{\left\lbrace{2}\right\rbrace}$$
b) $$\displaystyle{L}{\left\lbrace{e}^{{{2}{t}}}\right\rbrace}$$
c) $$\displaystyle{L}{\left[{e}^{{-{3}{t}}}\right]}$$

The function
$$\begin{cases}t & 0\leq t<1\\ e^t & t\geq1 \end{cases}$$
has the following Laplace transform,
$$L(f(t))=\int_0^1te^{-st}dt+\int_1^\infty e^{-(s+1)t}dt$$
True or False

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\)
To determine the solution of the initial value problem
$${y}{''}+{4}{y}= \sin{{t}}+{u}_{\pi}{\left({t}\right)} \sin{{\left({t}-\pi\right)}}:$$
$$y(0) = 0,$$
$$y'(0) = 0.$$
Also, draw the graphs of the solution and of the forcing function and explain the relation between the solution and the forcing function..
A function f (t) that has the given Laplace transform F (s).
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 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, $$L^{-1}[F(s)] = f(t)$$
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