Question

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

Laplace transform

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

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 $$L\left\{f(t)\right\}=\frac{6(s+4)}{(s^2+8s+25)^2}$$