Find the Laplace transform Lleft{u_3(t)(t^2-5t+6)right} a) F(s)=e^{-3s}left(frac{2}{s^4}-frac{5}{s^3}+frac{6}{s^2}right) b) F(s)=e^{-3s}left(frac{2}{s^3}-frac{5}{s^2}+frac{6}{s}right) c) F(s)=e^{-3s}frac{2+s}{s^4} d) F(s)=e^{-3s}frac{2+s}{s^3} e) F(s)=e^{-3s}frac{2-11s+30s^2}{s^3}

Question
Laplace transform
asked 2020-11-02
Find the Laplace transform \(L\left\{u_3(t)(t^2-5t+6)\right\}\)
\(a) F(s)=e^{-3s}\left(\frac{2}{s^4}-\frac{5}{s^3}+\frac{6}{s^2}\right)\)
\(b) F(s)=e^{-3s}\left(\frac{2}{s^3}-\frac{5}{s^2}+\frac{6}{s}\right)\)
\(c) F(s)=e^{-3s}\frac{2+s}{s^4}\)
\(d) F(s)=e^{-3s}\frac{2+s}{s^3}\)
\(e) F(s)=e^{-3s}\frac{2-11s+30s^2}{s^3}\)

Answers (1)

2020-11-03
Step 1
According to the given information, it is required to determine the laplace transform of the given function.
\(L\left\{u_3(t)(t^2-5t+6)\right\}\)
Step 2
Use the below formula to find the laplace transform.
\(L(u_c(t)g(t))=e^{-cs}L(g(t+c))\)
\(L(u_c(t))=\frac{e^{-cs}}{s}\)
Step 3
Solving further to get:
\(L(u_3(t)(t^2-5t+6))=L(u_3(t)t^2)-5L(u_3(t)t)+6L(u_3(t))\)
\(=e^{-3s}L((t+3)^2)-5e^{-3s}L(t+3)+\frac{6e^{-3s}}{s}\)
\(=e^{-3s}\left[\frac{2}{s^3}+\frac{6}{s^2}+\frac{9}{s}\right]-5e^{-3s}\left[\frac{1}{s^2}+\frac{3}{s}\right]+\frac{6e^{-3s}}{s}\)
\(=e^{-3s}\left[\frac{2}{s^3}+\frac{6}{s^2}+\frac{9}{s}-\frac{5}{s^2}-\frac{15}{s}+\frac{6}{s}\right]\)
\(=e^{-3s}\left[\frac{2}{s^3}+\frac{1}{s^2}\right]\)
\(L(u_3(t)(t^2-5t+6))=e^{-3s}\left[\frac{2+s}{s^3}\right]\)
Step 4
Therefore, the correct option is (D).
0

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