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 $\mathcal{L}\{t^2\cdot \sinh(2t)\}$ by differentiating $f(t)$.
Solution:
$$\mathcal{L}\{t\cdot f(t)\}=-F'(s)$$
$$\mathcal{L}\{t^2\cdot f(t)\}=F''(s)$$
$$f(t)=\sinh(2t)$$
$$F(s)=\frac{2}{s^2-4}$$
$$F'(s)=\frac{-1\cdot 2s\cdot 2}{(s^2-4)^2}=-\frac{4s}{(s^2-4)^2}$$
$$F''(s)=-\frac{4}{(s^2-4)^2}-\frac{-2\cdot 2s\cdot4s}{(s^2-4)^3}=-\frac{4}{(s^2-4)^2}+\frac{16s^2}{(s^2-4)^3}$$
Answer:
$$\mathcal{L}\{t^2\cdot \sinh(2t)\}=F''(s)=-\frac{4}{(s^2-4)^2}+\frac{16s^2}{(s^2-4)^3}$$
I also claim that for even powers of $t^n$ there is no need change the sign of the derivative because they cancel out. Is this a good assumption?
To understand what happens with the derivatives, let's look at the definition of the Laplace transform of a generic function of $t^n\sinh(bt)$. $$ \mathcal{L}\{t^n\sinh(bt)\} = \int_0^{\infty}t^n\sinh(bt)e^{-st}dt = \frac{\partial^n}{\partial s^n}(-1)^n\int_0^{\infty}\sinh(bt)e^{-st}dt = \frac{\partial^n}{\partial s^n}(-1)^n\mathcal{L}\{\sinh(bt)\} $$ Since we are taking the derivative of $e^{-st}$, we pick up a negative for every derivative. Now, as you claim, for even derivatives, we have a positive. That is due to $(-1)(-1)\cdots (-1)$ an even number of times which gives back $1$. Therefore, for odd derivatives, we would pick up a negative.