I have the following Laplace transform problem:
Prove the following change of scale result:
$$\mathcal{L} \{ F(at) \} = \dfrac{1}{a} f \left( \dfrac{s}{a} \right).$$
Hence evaluate the Laplace transforms of the two functions
(a) $t \cos(6t)$, (b) $t^2 \cos(7t)$.
The solution is as follows:
The proof proceeds by using the definition as follows:
$$\mathcal{L} \{ F(at) \} = \int_0^\infty e^{-st} F(at) \ dt = \int_0^\infty e^{-su/a} F(u) \ du/a,$$
which gives the result. Evaluation of the two Laplace transforms follows from using the results of Exercise 5 alongside the change of scale result just derived, for (a) $a = 6$ and (b) $a = 7$. The answers are
(a) $\dfrac{-36 + s^2}{(36 + s^2)^2}$, (b) $\dfrac{2s(s^2 - 147)}{(s^2 + 49)^3}$.
The aforementioned "results of exercise 5" are as follows:
$$\mathcal{L}\{ t \cos(t) \} = \dfrac{-1 + s^2}{(1 + s^2)^2},$$
$$\mathcal{L}\{ t^2 \cos(t) \} = \dfrac{2s^3 - 6s}{(1 + s^2)^3}$$
I don't understand how the provided information is sufficient for calculating the two Laplace transforms. In particular, I don't see how the results of exercise 5 are supposed to be used. Or am I misunderstanding this, and we are actually supposed to solve for the two Laplace transforms using just $\int_0^\infty e^{-su/a} F(u) \ du/a$? The solution seems to imply that we don't need to solve for these from the definition of Laplace transforms, and can instead use some kind of shortcut by knowing the results of exercise 5, but this isn't clear.
I would greatly appreciate it if people would please take the time to clarify this.
The proof shows the following: If $F$ has Laplace transform $f$, then the function $$ G(t):=F(at) $$ has Laplace transform $g$, where $g$ is defined by $$ g(s):=\frac1a f\left(\frac sa\right).\tag{*} $$ So for example, given that the LT of $F(t):=t \cos t$ is $$f(s):=\frac{-1+s^2}{(1+s^2)^2}\tag1$$ the assertion is that the LT of $G(t):=6t \cos (6t)$ [note the 6 in front!] is $$ g(s):=\frac16 f\left(\frac s6\right).\tag2 $$ Evaluate $g$ by taking (1), plugging in $s/6$ in place of $s$, dividing the result by $6$, and simplifying. Finally obtain the LT of $t\cos(6t)$ by dividing (2) by $6$ again. You will get the advertised result $g(s)=\dfrac{-36 + s^2}{(36 + s^2)^2}$.