Question

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

Laplace transform
ANSWERED
asked 2021-03-05
Find the laplace transform of the following:
\(a) t^2 \sin kt\)
\(b) t\sin kt\)

Answers (1)

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