Find the inverse laplace transform of frac{6s+9}{s^2+17} s>0 y(t)=dots

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Find the inverse laplace transform of frac{6s+9}{s^2+17} s>0 y(t)=dots

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asked 2020-10-23

Find the inverse Laplace transform of
\(\frac{6s+15}{(s^2+25)}\)
\(s>0\)

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asked 2021-02-16

Solution of I.V.P for harmonic oscillator with driving force is given by Inverse Laplace transform
\(y"+\omega^{2}y=\sin \gamma t , y(0)=0,y'(0)=0\)
1) \(y(t)=L^{-1}\bigg(\frac{\gamma}{(s^{2}+\omega^{2})^{2}}\bigg)\)
2) \(y(t)=L^{-1}\bigg(\frac{\gamma}{s^{2}+\omega^{2}}\bigg)\)
3) \(y(t)=L^{-1}\bigg(\frac{\gamma}{(s^{2}+\gamma^{2})^{2}}\bigg)\)
4) \( y(t)=L^{-1}\bigg(\frac{\gamma}{(s^{2}+\gamma^{2})(s^{2}+\omega^{2})}\bigg)\)

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asked 2021-02-08

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) \(\cos(t)+4\sin(t)\)

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asked 2020-10-25

Please provide steps
The inverse Laplace transform for
\(\displaystyle{F}{\left({s}\right)}=\frac{8}{{{s}+{9}}}-\frac{6}{{{s}^{2}-\sqrt{{3}}}}\) is
a) \(\displaystyle{8}{e}^{{-{9}{t}}}-{6} \sin{{h}}{{\left({3}{t}\right)}}\)
b) \(\displaystyle{8}{e}^{{-{9}{t}}}-{6} \cos{{h}}{\left({3}{t}\right)}\)
c) \(\displaystyle{8}{e}^{{{9}{t}}}-{6} \sin{{h}}{\left({3}{t}\right)}\)
d) \(\displaystyle{8}{e}^{{{9}{t}}}-{6} \cos{{h}}{\left({3}{t}\right)}\)

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asked 2021-02-20

Find the inverse Laplace transform of (any two)
i) \(\frac{(s^2+3)}{s(s^2+9)}\)
ii) \(\log\left(\frac{(s+1)}{(s-1)}\right)\)

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asked 2020-11-01

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.

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asked 2021-02-25

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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asked 2020-11-05

Find the inverse laplace transform of the function
\(Y(s)=\frac{e^{-s}}{s(2s-1)}\)

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asked 2021-02-09

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) - ?

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asked 2021-01-15

Let \(y(t)=\int_0^tf(t)dt\) If the Laplace transform of y(t) is given \(Y(s)=\frac{19}{(s^2+25)}\) , find f(t)
a) \(f(t)=19 \sin(5t)\)
b) none
c) \(f(t)=6 \sin(2t)\)
d) \(f(t)=20 \cos(6t)\)
e) \(f(t)=19 \cos(5t)\)

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Answer 1

Step 1
Given the Laplace transform of a function y(t) such that
\(L\left\{y(t)\right\}=\frac{6s+9}{s^2+17}\)
Find f(t) using inverse Laplace transform.
Step 2
First, separate the numerator.
\(\frac{6s+9}{(s^2+17)}=6 \frac{s}{(s^2+17)}+9 \frac{1}{(s^2+17)}\)
Since \(L^{-1}\left\{\frac{a}{(s^2+a^2)}\right\}=\sin (at)\) , multiply the numerator and denominator on the second sum by \(\sqrt{17}\)
\(\frac{6s+9}{(s^2+17)}=6 \frac{s}{(s^2+(\sqrt{17})^2}+\frac{9}{(\sqrt{17})} \cdot \frac{\sqrt{17}}{s^2+(\sqrt{17})^2}\)
Step 3
Now, take inverse Laplace transform on both sides.
\(L^-1\left\{\frac{6s+9}{(s^2+17)}\right\}=6L^ {-1}\left\{\frac{s}{s^2+(\sqrt{17})^2}\right\}+\frac{9}{\sqrt{17}}L^-1\left\{\frac{\sqrt{17}}{s^2+(\sqrt{17})^2}\right\}\)
Since \(L^-1\left\{\frac{a}{(s^2+a^2)}\right\}=\sin (at) \text{ and } L^-1\left\{s/(s^2+a^2)\right\}=\cos(at)\)
\(L^-1\left\{\frac{6s+9}{(s^2+17)}\right\}=6\cos(\sqrt{17t})+9/(\sqrt{17})sin(\sqrt{17t})\)
Therefore, \(y(t)=6\cos(\sqrt{17t})+\frac{9}{(\sqrt{17})}\sin(\sqrt{17t})\)

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Find the inverse laplace transform of frac{6s+9}{s^2+17} s>0 y(t)=dots

Find the inverse laplace transform of frac{6s+9}{s^2+17} s>0 y(t)=dots

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