Find the laplace transform of the following: MULTIPLICATION BY POWER OF t g(t)=(t^2-3t+2)sin(3t)

Question
Laplace transform
asked 2020-12-24
Find the laplace transform of the following:
MULTIPLICATION BY POWER OF t
\(g(t)=(t^2-3t+2)\sin(3t)\)

Answers (1)

2020-12-25
Step 1
To find Laplace transform of \(g(t)=(t^2-3t+2)\sin(3t)\), we simplify the expression and we get
\(g(t)=t^2\sin(3t)-3t\sin(3t)+2\sin(3t)\)
Using the following results:
\(1. L(f(t))=F(s)\)
\(2. L(t^nf(t))=(-1)^n\frac{d^nF(s)}{ds^n}\)
\(3. L(\sin(at))=\frac{a}{s^2+a^2}\)
Step 2
Using the above results and finding the required transforms,
\(L(\sin(3t))=\frac{3}{s^2+3^2}=\frac{3}{s^2+9}\)
\(L(t^2\sin(3t))=(-1)\frac{2}{d}\)
Now,
\(L(-3t \sin(3t))=-3(-1)\frac{d}{ds}(L(\sin(3t)))\)
\(=3\frac{d}{ds}\left(\frac{3}{s^2+9}\right)\)
\(=3\left(\frac{-3 \cdot 2s}{(s^2+9)^2}\right)\)
\(=\frac{-18s}{(s^2+9)^2}\)
Now,
\(L(2\sin(3t))=2\frac{3}{(s^2+9)}\)
\(=\frac{6}{(s^2+9)}\)
Step 3
Therefore,
\(L(g(t))=L(t^2\sin(3t)-3t\sin(3t)+2\sin(3t))\)
\(=L[t^2\sin(3t)]+L[-3t\sin(3t)]+L[2\sin(3t)]\)
Substituting the values of Laplace transforms from above step, we get
\(G(s)=\frac{6[3s^2-9]}{(s^2+9)^3}-\frac{18s}{(s^2+9)^2}+\frac{6}{(s^2+9)}\)
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