Question

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

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

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