What is the derivation of 2e^{2t}\sin{t} in Laplace transform?

[faust9]

OK, here's the problem

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

What I did:
$$L^{-1}\{\frac{s}{(s+2)^2+1}\}$$

$$=e^{2t}\cos{t}$$

The book says:
$$=e^{2t}\cos{t}-2e^{2t}\sin{t}$$


Where did the $2e^{2t}\sin{t}$ come from?

Mahalo.

 

[cookiemonster]

Mathematica is saying that it's neither of those...

cookiemonster

 

[turin]

The cannonical form has an (s+2) in the numerator, not just s (so your invesrse transform calculation was actually incorrect). In order to account for that, you have to add another rational function that has a numerator of -2, which just happens to be (-2) times the cannonical form with sin instead of cos.

 

[faust9]

Thanks a lot. I missed about a month of Diff Eq due to illness and an auto accident so I've been learning it all on my own. I appreciate the Q&A forum here and all those who help out.