An inverse Laplace transform

Homework Statement

.......find the inverse transform of the given function.

F(s)= (2s+2)/(s2+4s+5)

Homework Equations

On another page, I see these 2:

f(t)=eatsin(bt)--------->F(s)=b/[(s-a)2+b2)
f(t)=eatcos(bt)--------->F(s)=(s-a)/[(s-a)2+b2)

The Attempt at a Solution

This is my first time doing anything with Laplace transforms.

I can change the denominator to (s+2)2 + 1 and factor a -2 out of the numerator. Then it looks like the relevant equations.

-2((s-1/2)/((s+2)2+1))

But I don't understand where to go from here. The -2, of course, is just a constant that can chill out front. Judging from the answer in the back of the book, I use a combination of both relevant equations. Explain the algebra involved here.

Dick
Homework Helper
Ok, so a=(-2) and b=1, right? The numerator of the first expression is 1 and the second is s+2. Looks like you want to solve c*1+d*(s+2)=(2s+2) for c and d. Then multiply your first f(t) by c and the second f(t) by d.

Ok, so a=(-2) and b=1, right? The numerator of the first expression is 1 and the second is s+2. Looks like you want to solve c*1+d*(s+2)=(2s+2) for c and d. Then multiply your first f(t) by c and the second f(t) by d.

Let's slow down for a second.

I have (s-1/2)/((s+2)2+1).

That isn't in the form b/((s-a)2+b2) because the b's are inconsistent. If I make a=-2, and b=1, the denominator works out, but then the numerator doesn't.

It also isn't in the form (s-a)/((s-a)2+b2) because the a's are inconsistent. If I do the same, the numerator doesn't work out.

Explain this to me as if I was a 5-year-old.

http://blog.oregonlive.com/petoftheday/2008/04/large_thanks.bmp [Broken]

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Dick