Question

# Find the laplace transform of the following: a) t^2 sin kt b) tsin kt

Laplace transform
Find the laplace transform of the following:
$$a) t^2 \sin kt$$
$$b) t\sin kt$$

2021-03-06
$$\text{Step 1}$$
$$\text{Recall the following.}$$
$$L\left\{t^kf(t)\right\}=(-1)^k\frac{d^k}{ds^k}(L\left\{f(t)\right\})$$
$$L\left\{\sin(at)\right\}=\frac{a}{s^2+a^2}$$
$$\text{Step 2}$$
$$\text{(a) Obtain the Laplace transform as follows.}$$
$$(a) f(t)=t^2\sin (kt)$$
$$L\left\{\sin(kt)\right\}=\frac{k}{s^2+k^2}$$
$$L\left\{t^2\sin(kt)\right\}=(-1)^2\frac{d^2}{ds^2}(L\left\{\sin(kt)\right\})$$
$$=\frac{d^2}{ds^2}\bigg(\frac{k}{s^2+k^2}\bigg)$$
$$=\frac{d}{ds}\bigg(\frac{d}{ds}\bigg(\frac{k}{s^2+k^2}\bigg)\bigg)$$
$$=\frac{d}{ds}\bigg(-\frac{2ks}{(s^2+k^2)^2}\bigg)$$
$$=-2k\frac{d}{ds}\bigg(\frac{s}{(s^2+k^2)^2}\bigg)$$
$$=-2k\frac{(s^2+k^2)^2\frac{d}{ds}(s)-s\frac{d}{ds}\left[(s^2+k^2)^2\right]}{((s^2+k^2)^2)^2}$$
$$=-2k\frac{(s^2+k^2)^2-4s^2\left[(s^2+k^2)\right]}{((s^2+k^2)^2)^2}$$
$$=-2k\frac{(s^2+k^2)-4s^2}{(s^2+k^2)^3}$$
$$=\frac{2k(3s^2-k^2)}{(s^2+k^2)^3}$$
$$\text{Step 3}$$
$$\text{(b) Obtain the Laplace transform as follows.}$$
$$(b) f(t)=t\sin(kt)$$
$$L\sin(kt)=\frac{k}{s^2+k^2}$$
$$L\left\{t\sin(kt)\right\}=(-1)^1\frac{d^1}{ds^1}(L\left\{\sin(kt)\right\})$$
$$=-\frac{d}{ds}\bigg(\frac{k}{s^2+k^2}\bigg)$$
$$=-k\frac{d}{ds}\bigg(\frac{1}{s^2+k^2}\bigg)$$
$$=-k\bigg(\frac{-1}{(s^2+k^2)^2}\bigg)\frac{d}{ds}(s^2+k^2)$$
$$=\frac{k}{(s^2+k^2)^2}(2s)$$
$$=\frac{2ks}{(s^2+k^2)^2}$$