Question

The Laplace transform of tu(t-2)=e^{-2s}left[frac{s+2s^2}{s^3}right] True or False?

Laplace transform
ANSWERED
asked 2021-02-03
The Laplace transform of \(tu(t-2)=e^{-2s}\left[\frac{s+2s^2}{s^3}\right]\)
True or False?

Answers (1)

2021-02-04
Step 1
The given function is \(tu(t-2)\)
To find the Laplace transform proceed as follows
\(L\left\{tu(t-2)\right\}\) . The function \(u(t-2)\) is unit step function .
Find the Laplace of \(u(t-2)\)
\(L\left\{u(t-2)\right\}=\frac{e^{-2s}}{s}\)
The function \(u(t-2)\) is multiplied by t, use the formula for the multiplication by t^n use the formula \(L\left\{f(t)\right\}= L\left\{f(t)\right\}=(\bar f)(s) \text{ then } L\left\{t^nf(t)\right\}=(-1)^n \frac{d^n}{ds^n}(\bar f)(s)\)
here n =1
therefore, \(L\left\{tu(t-2)\right\}=(-1)\frac{d}{ds}\frac{e^{-2s}}{s}\)
Step 2
Solve \(L\left\{tu(t-2)\right\}=(-1)\frac{d}{ds}\frac{e^{-2s}}{s}\)
\(L\left\{tu(t-2)\right\}=(-1)\frac{\frac{sd}{ds}e^{-2s}-e^{-2s}\frac{d}{ds}s}{s^2}\)
\(=(-1)\left(\frac{s(-2)e^{-2s}-e^{-2s}(1)}{s^2}\right)\)
\(=\left(\frac{2se^{-2s}+e^{-2s}}{s^2}\right)\)
\(=e^{-2s}\left(\frac{2s+1}{s^2}\right) \cdot \frac{s}{s}\)
\(=e^{-2s}\left(\frac{2s^2+s}{s^3}\right)\)
The Laplace transform of the given function is \(e^{-2s}\left(\frac{2s^2+s}{s^3}\right)\)
The answer is TRUE
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