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 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)

Question
Laplace transform
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, \(L^{-1}[F(s)] = f(t)\)

Answers (1)

2021-02-26
Step 1
We use the results from table of Laplace transform.
\(a) f(t)=(t+5)^2+t^2e^{5t} = t^2+10t+25+t^2e^{5t}\)
\(\Rightarrow L\left\{f(t)\right\}=L\left\{t^2+10t+25+t^2e^{5t}\right\}=L\left\{t^2\right\}+10L\left\{t\right\}+L\left\{25\right\}+L\left\{t^2e^{5t}\right\}\)
\(F(s)=\frac{2}{s^3}+\frac{10}{s^2}+\frac{25}{s}+\frac{2}{(s-5)^3}\)
\(b) f(t)=2e^{-t}\cos\left(3t+\frac{\pi}{4}\right) \Rightarrow 2e^{-t}\left[\cos(3t) \cdot \cos\left(\frac{\pi}{4}\right) - \sin(3t) \cdot \sin\left(\frac{\pi}{4}\right)\right]\)
\(\Rightarrow 2e^{-t}\cos(3t) \cdot \cos\left(\frac{\pi}{4}\right) - 2e^{-t}\sin(3t) \cdot \sin\left(\frac{\pi}{4}\right)\)
\(\Rightarrow L\left\{f(t)\right\}=2\cos\left(\frac{\pi}{4}\right)L\left\{e^{-t}\cos(3t)\right\}-2\sin\left(\frac{\pi}{4}\right)L\left\{e^{-t}\sin(t)\right\}\)
we take constant term Laplace
\(\Rightarrow F(s)=\frac{\sqrt2s}{(s+1)^2+9} - \frac{2}{(s+1)^2+9}\)
\(c) F\left(\frac{7s^2-37s+64}{s(s^2-8s+16)}\right)=\frac{4}{s}+\frac{3}{(s-4)}+\frac{7}{(s-4)^2}\)
(by partial fraction)
Then \(L^{-1}\left\{F(s)\right\}=L^{-1}\left\{\frac{4}{s}\right\}+L^{-1}\left\{\frac{3}{(s-4)}\right\}+L^{-1}\left\{\frac{7}{(s-4)^2}\right\}\)
\(f(t)=4+3e^{4t}+7te^{4t}\)
\(d) F(s)=e^{-7s}\left(\frac{1}{s}+\frac{s}{(s^2+1)}\right)=\frac{e^{-7s}}{s} + \frac{s e^{-7s}}{(s^2+7)}\)
\(\Rightarrow L^{-1}\left\{F(s)\right\}=L^{-1}\left\{\frac{e^{-7s}}{s}\right\}+L^{-1}\left\{\frac{s e^{-7s}}{(s^2+7)}\right\}\)
\(=u(t-7)+u(t-7)\cos(t-7)\)
All the results teple of Laplace transform
0

Relevant Questions

asked 2021-02-08
Find the inverse Laplace transform of \(F(s)=\frac{(s+4)}{(s^2+9)}\)
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b)non of the above
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asked 2021-05-05

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Previous studies show that \( \sigma_1 = 19 \).
For Englewood (a suburb of Denver), a random sample of \( n_2 = 12 \) winter days gave a sample mean pollution index of \( x_2 = 37 \).
Previous studies show that \( \sigma_2 = 13 \).
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\( H_0:\mu_1=\mu_2.\mu_1>\mu_2 \)
\( H_0:\mu_1<\mu_2.\mu_1=\mu_2 \)
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(b) What sampling distribution will you use? What assumptions are you making? NKS The Student's t. We assume that both population distributions are approximately normal with known standard deviations.
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.
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Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, we can not say that the mean population pollution index for Englewood is different than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains only negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is less than that of Denver.
asked 2021-01-15
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True or False

asked 2020-12-30
The Laplace inverse of \(L^{-1}\left[\frac{s}{s^2+5^2}\right]\) is
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Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.
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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.
asked 2020-11-02
Find the Laplace transform \(L\left\{u_3(t)(t^2-5t+6)\right\}\)
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Please provide steps
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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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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)
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b) \(X(s)=\frac{1}{(2s^2+8s+20)(s^2+2s+2)(s+8)}\)
c) \(X(s)=\frac{1}{s^2(s^2+2s+5)(s+3)}\)
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