Laplace transform of $t^2 \cos \omega t$

Question

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:

$$\frac{2s(s^2 - 3 \omega^2)}{(s^2 + \omega^2)^3}$$

Answer 1

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 $$\frac{2s(s^2 - 3 \omega^2)}{(s^2 + \omega^2)^3}$$