I have to take the laplace transform of $te^{3t}\cos(3t)$ So I started out with the property $L\{t^nf(t)\} = (-1)^n\frac{d}{ds} L\{f(t)\}$ Which yielded $-s^2-b^2+2s^2/(s^2+b^2)^2$. If I did it correctly.
Then I would be using $L\{e^{at} f(t)\} = L\{f\}(s-a)$
When I plug in $s, a, b,$ I get in the end $-3(s-3)^2+49/((s-3)^2+49)^2$
The denominator is correct, but the book says the numerator is $(s-3)^2-49.$ What did I do wrong?
$F(s) = \mathcal{L}\{te^{3t}\cos3t\} $
$F(s) = \mathcal{L}\{t\cos3t\}_{s\to s-3}$
$F(s) = -\frac{d}{ds}\mathcal{L}\{cos3t\}_{s\to s-3}$
$F(s) = -\frac{d}{ds}\{\frac{s}{s^2+9}\}_{s\to s-3}$
$F(s) = -\bigg[\frac{1\cdot(s^2+9) - s\cdot(2s)}{(s^2+9)^2}\bigg]_{s\to s-3}$
$F(s) = -\bigg[\frac{-s^2 + 9}{(s^2+9)^2}\bigg]_{s\to s-3} = \frac{(s-3)^2-9}{((s-3)^2 + 9)^2}$
$$F(s) = \frac{s^2 - 6s}{(s^2-6s+18)^2}$$