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Find f(t) given an inverse Laplace transform

JJBladester

Gold Member
286
2
1. Homework Statement

Find [tex]f(t)[/tex].

2. Homework Equations

[tex]L^{-1}\left\{\frac{s}{s^{2}+4s+5}\right\}[/tex]

3. The Attempt at a Solution

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

[tex]L^{-1}\left\{\frac{s}{s^{2}+4s+5}\right\}[/tex] =

[tex]L^{-1}\left\{\frac{s}{(s+2)^{2}+1}\right\}[/tex]

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

[tex]coskt = L^{-1}\left\{\frac{s}{(s^{2}+k^{2})}\right\}[/tex]

[tex]L\left\{e^{at}f(t)\right\} = F(s-a)}[/tex]

with [tex]a[/tex] being -2 and [tex]k[/tex] being 1 to get an answer of:

[tex]e^{-2t}cos(t)[/tex]

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

[tex]e^{-2t}cos(t)-2e^{-2t}sin(t)[/tex]

My question is where does the [tex]-2e^{-2t}sin(t)[/tex] come from?
 

djeitnstine

Gold Member
614
0
Are you sure that is all of the Laplace transform?
 

rock.freak667

Homework Helper
6,230
31
[tex]\frac{s}{(s+2)^2+1}=\frac{s+2 -2}{(s+2)^2+1}[/tex]

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

Gold Member
614
0
Good call, I knew that was missing but for some reason failed to mention it =(
 

JJBladester

Gold Member
286
2
[tex]\frac{s}{(s+2)^2+1}=\frac{s+2 -2}{(s+2)^2+1}[/tex]

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.
 

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