Laplace Transform of $100e^{-5t}\sin10t$

Question

Can anybody help me with the answer of this question?

$$100e^{-5t}\sin10t$$

Answer 1

We have:

$$\mathcal{L}(\sin 10 t) = \dfrac{10}{s^2 + 100}$$

We have:

$$\mathcal{L}(e^{-5 t} \sin 10 t) = F(s+5) = \dfrac{10}{(s+5)^2 + 100}$$

Lastly, we multiply by $100$, to yield:

$$\mathcal{L}(100e^{-5t}\sin10t) = \dfrac{1000}{(s+5)^2 + 100}$$

Answer 2

Use the first shifting theorem:

$$\mathcal L(e^{-at}f(t))=F(s+a)$$

Where $F(s)$ is the Laplace Transform of f(t) , so knowing that: $$\mathcal L( a \sin(wt))= a\mathcal L(\sin(wt))=a\frac{w}{s^2+w^2}$$

$$f(t)=100\sin(10t) \implies F(s)=100*\frac{10}{(s^2+10^2)}$$

$$\mathcal L(e^{-5t}f(t))=\frac{1000}{((s+5)^2+100)}$$