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.