Question

Find Laplace transform of the given functionte^{-4t}sin 3t

Laplace transform
ANSWERED
asked 2020-11-06

Find Laplace transform of the given function
\(te^{-4t}\sin 3t\)

Expert Answers (1)

2020-11-07
Step 1
Concept Used:
\(L\left\{t^kf(t)\right\}=(-1)k(d^k)\frac{L\left\{f(t)\right\}}{d}s^k\)
also,
\(L\left\{\sin(at)\right\}=\frac{a}{(s^2+a^2)}\)
\(L\left\{e^{-bt}\sin(at)\right\}=\frac{a}{(s+b)^2+a^2}\)
Step 2
Given: Given function is \(f(t)=te^{-4t}\sin(3t)\)
We want to calculate Laplace transform
Calculation:
Given function is \(f(t)=te^{-4t}\sin(3t)\)
Therefore,
\(L\left\{f(t)\right\}=L\left\{te^{-4t}\sin(3t)\right\}\)
\(=(-1)1d\frac{[L\left\{e^{-4t}\sin(3t)\right\}]}{ds}\)
\(=-\frac{d}{ds}\left[\frac{3}{(s+4)^2+9}\right]\)
\(=-\frac{d}{ds}\left[\frac{3}{s^2+8s+16+9}\right]\)
\(=-\frac{d}{ds}\left[\frac{3}{s^2+8s+25}\right]\)
\(=-3\frac{d(s^2+8s+25)^{-1}}{ds}\)
\(=-3(-1)(s^2+8s+25)^{-2}\frac{d(s^2+8s+25)}{ds}\)
\(\left\{\text{By chain rule }\right\}\)
\(=3(s^2+8s+25)^-2(2s+8)\)
\(L\left\{f(t)\right\}=\frac{6(s+4)}{(s^2+8s+25)^2}\)
Step 3
Answer: The Laplace transform of \(f(t)=te^{-4t}\sin(3t)\) is \(L\left\{f(t)\right\}=\frac{6(s+4)}{(s^2+8s+25)^2}\)
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