Calculate Laplace transform of the product of t and f(t) by differenitating f(t) (5.5-4)

Question

Request:

Please check my work. State where errors, if any, occurred and how to correct them. Is there a better way to calculate the transform other than the present method given?

Given:

Find the Laplace transform $\mathcal{L}\{t\cdot \sin(3t)\}$ by differentiating $f(t)$.

Solution:

$$\mathcal{L}\{t\cdot f(t)\}=-F'(s)$$

$$f(t)=\sin(3t)$$

$$F(s)=\frac{s}{s^2+9}$$

$$F'(s)=\frac{s\cdot(-1)\cdot2s}{(s^2+9)^2}+\frac{1}{(s^2+9)}=-\frac{2s^2}{(s^2+9)^2}+\frac{1}{(s^2+9)}$$

$$\mathcal{L}\{t\cdot \sin(3t)\}=-F'(s)=\frac{2s^2}{(s^2+9)^2}-\frac{1}{(s^2+9)}$$

Answer 1

You are considering the Laplace transform of $\cos(3t)$ while you should consider the Laplace transform of $\sin(3t)$

$$ F(s) = \frac{3}{s^2+9}. $$