Find inverse laplace of the following?

Question

Find the inverse Laplace of $$\frac{e^{-2s} - 3e^{-4s}}{s+2}.$$

I tried splitting up the function into two, so I get $$\frac{e^{-2s}}{s+2} - \frac{3e^{-4s}}{s+2}.$$ Then we have $$e^{-2s}(s+2)^{-1} - e^{-4s}(\frac{3}{s+2}).$$

Taking $f(t)$, we have $f(t) = 4e^{-2t}$. I am stuck here, though. What do I do?

Answer 1

Using the Laplace's Frequency Translation Property:

$$ L^{-1} \{e^{-as}\} = H(t-a)f(t-a)$$

with $H(t)$ the heavyside function, or step function. This function points out that everything before the point $t=a$ is zero.

So

$$ F(s)e^{-2s} \to H(t-2)f(t-2) $$ $$ F(s)e^{-4s} \to H(t-4)f(t-4) $$

And remembering that: $$ L\{\frac{1}{s+2}-\frac{3}{(s+2)}\}=e^{-2t}-3e^{-2t}$$

We have just to put them together, so the solution is:

$$H(t-2)e^{-2(t-2)}-3H(t-2)e^{-2(t-4)} $$

Answer 2

Using the Properties of Laplace transform, $$\text{as }L\{e^{bt}\}=\frac1{s-b}\implies L^{-1}\left(\frac1{s-b}\right)=e^{bt}\implies L^{-1}\left(\frac1{s+2}\right)=e^{-2t}$$

Let $F(S)=\frac1{s+2}$ and $f(t)=e^{-2t}$

$$\text{Now, }L^{-1}\left(e^{-as}F(s)\right)=f(t-a)u(t-a)$$ where $u(t)$ is the Heaviside step function