Find f(t) given an inverse Laplace transform

[JJBladester]

Homework Statement

Find f(t).

Homework Equations

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

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?

 

[djeitnstine]

Are you sure that is all of the Laplace transform?

 

[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.

 

[djeitnstine]

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

 

[JJBladester]

\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.

Aaaaha! That looks like a Calc II trick coming back to light. The chapter in my Differential Equations book dealing with Laplace Transforms does talk a lot about fancy algebra techniques like partial fraction decomposition, rearranging constants, and completing the square. I just forgot this one in particular.

Thanks for the explanation.