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

\(G(s)=\frac{6[3s^2-9]}{(s^2+9)^3}-\frac{18s}{(s^2+9)^2}+\frac{6}{(s^2+9)}\)

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

\(G(s)=\frac{6[3s^2-9]}{(s^2+9)^3}-\frac{18s}{(s^2+9)^2}+\frac{6}{(s^2+9)}\)