Inverse Laplace transform of 5/(s+1)·exp(-2s)

[daf1]

I want to find the following inverse Laplace transform:

[5/(s+1)]exp(-2s)

I was thinking of using the second shift theorem to no avail.

I have tried the question and my answer is :
5(exp-t)u(t-2)

This does not seem to be a plausible answer since I modified the second
shift theorem.

{If only the question was to inverse [5/(s+2)]exp(-2s),
then my answer would have been 5(exp-2t)u(t-2) (which hopefully is right).
However the question is different.}

 

[EugP]

I just want to make sure I understand the equation. Is it:

$$\frac{5e^{-2s}}{s+1}$$ ?

If it is, I think this is simpler than you're making it.

$$\frac{5e^{-2s}}{s+1} = \frac{K}{s+1}$$

Multiply both sides by $$s+1$$ and evaluate at $$s = -1$$:

$$5e^{-2(-1)} = K$$

$$5e^{2} = K$$

Now plug it back in:

$$\frac{5e^{2}}{s+1}$$

Now the inverse laplace would be:

$$L^{-1} = 5e^{2}e^{-t} u(t)$$

i.e.

$$L^{-1} = 5e^{2-t} u(t)$$

I hope this helped.