Laplace transform of $t^2 \cos \omega t$ I have to find the Laplace transform of: $$f(t) = t^2 \cos \omega t$$
We know that,
$$\mathcal{L} (t^n f(t)) = (-1)^n \frac{d^n}{ds^n}{F(s)}$$
so , $$\mathcal{L}(\cos \omega t) = \frac{s}{s^2 + \omega^2}$$
$$\therefore \mathcal{L} (t^2 \cos \omega t) = \frac{d^2}{ds^2} \frac{s}{s^2 + \omega^2}$$
let $$\frac{d^2}{ds^2} = F^{\prime\prime}(s)$$
so,
Using quotient rule of differentiation
$$F^\prime(s) = \frac{\omega^2 -s^2}{(s^2 + \omega^2)^2}$$
however, when I try to differentiate $F^\prime(s)$ again to find $F^{\prime\prime}(s)$ it is an entire mess. Can someone help me to differentiate it or is there any other way to find its Laplace Transformation.
The answer is:
$$\frac{2s(s^2 - 3 \omega^2)}{(s^2 + \omega^2)^3}$$
 A: OP, your process is correct and the only thing left is to find $F^{\prime\prime}(s)$.
$$F^\prime(s) = \frac{\omega^2 -s^2}{(s^2 + \omega^2)^2}$$
$$F^{\prime\prime}(s) = \frac{d}{ds} \frac{\omega^2 -s^2}{(s^2 + \omega^2)^2}$$
Quotient Rule:
$$\frac{d}{ds} \frac{f(s)}{g(s)} = \frac{f'(s)g(s)-g'(s)f(s)}{g(s)^2}$$
So,$$\frac{d}{ds} \left(\frac{\omega^2 -s^2}{(s^2 + \omega^2)^2}\right) =  \frac{(\frac{d}{ds}(\omega^2 -s^2))\ (s^2 + \omega^2)^2\ \ -\ \ (\frac{d}{ds}(s^2 + \omega^2)^2)\ (\omega^2 -s^2)}{((s^2 + \omega^2)^2)^2}$$
$$=\ \frac{(-2s)(s^2+\omega^2)^2\ -\ (2(s^2+\omega^2)\ (2s))\ (\omega^2 -s^2)}{(s^2 + \omega^2)^4}$$
$$=\ \frac{(-2s)(s^2+\omega^2)\ -\ (4s)\ (\omega^2 -s^2)}{(s^2 + \omega^2)^3}$$
$$=\ \frac{(-2s)(s^2+\omega^2)\ +\ (4s)\ (s^2-\omega^2)}{(s^2 + \omega^2)^3}$$
$$=\ \frac{(2s)\left(2(s^2-\omega^2)-(s^2+\omega^2)\right)}{(s^2 + \omega^2)^3}$$
$$=\ \frac{2s(s^2 - 3 \omega^2)}{(s^2 + \omega^2)^3}$$
$$\therefore\ F^{\prime\prime}(s) = \frac{d}{ds} \left(\frac{\omega^2 -s^2}{(s^2 + \omega^2)^2}\right)\ =\ \frac{2s(s^2 - 3 \omega^2)}{(s^2 + \omega^2)^3}$$
Thus, the Laplace transform of $f(t) = t^2 \cos \omega t$ is
$$\frac{2s(s^2 - 3 \omega^2)}{(s^2 + \omega^2)^3}$$
