# Find f(t) given an inverse Laplace transform

1. Mar 22, 2009

1. The problem statement, all variables and given/known data

Find $$f(t)$$.

2. Relevant equations

$$L^{-1}\left\{\frac{s}{s^{2}+4s+5}\right\}$$

3. The attempt at a solution

I tried completing the square to get to the solution and I ended up with:

$$L^{-1}\left\{\frac{s}{s^{2}+4s+5}\right\}$$ =

$$L^{-1}\left\{\frac{s}{(s+2)^{2}+1}\right\}$$

Then, I used the inverse of a transform for cosine and the first translation theorum:

$$coskt = L^{-1}\left\{\frac{s}{(s^{2}+k^{2})}\right\}$$

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

with $$a$$ being -2 and $$k$$ being 1 to get an answer of:

$$e^{-2t}cos(t)$$

However, I was wrong. The book had the following answer:

$$e^{-2t}cos(t)-2e^{-2t}sin(t)$$

My question is where does the $$-2e^{-2t}sin(t)$$ come from?

2. Mar 22, 2009

### djeitnstine

Are you sure that is all of the Laplace transform?

3. Mar 22, 2009

### rock.freak667

$$\frac{s}{(s+2)^2+1}=\frac{s+2 -2}{(s+2)^2+1}$$

now when you split that into two fractions, the one with s+2 in the numerator will give the cos term and the other will give the sin term, assuming I remember laplace transform correctly.

4. Mar 23, 2009

### djeitnstine

Good call, I knew that was missing but for some reason failed to mention it =(

5. Mar 23, 2009