A challenge Question on Inverse Laplace

[Theelectricchild]

Hey everyone, I am trying to solve this challenge question on finding the inverse laplace transform of

$$f(s)=\frac{5se^{-3s} - e^{-3s}}{s^{2}-4s+17}$$

Heres my reasoning: I feel that I should split up that denominator into two linear factors, however its that odd 17 that is very troublesome. In addition, I attempted to factor out an $$e^{-3s}$$ in the numerator and try to proceed from there--- but I tried to get help from the homework section but it seems like not as many posts come in there, and Id think avid diff eq experts hang out here more, so any help would be greatly appreciated in solving this problem! THANKS!

 

[HallsofIvy]

You CAN'T factor s²- 4s+ 17 since the equation s²- 4s+ 17= 0 does not have real roots but you can "complete the square":

s²- 4s+ 17= s²+ 4s+ 4+ 13= (s+2)²+ 13.

And yes, factoring the e^-3s out is a good idea.

 

[Tide]

You CAN'T factor s²- 4s+ 17 since the equation s²- 4s+ 17= 0 does not have real roots but you can "complete the square":

s²- 4s+ 17= s²+ 4s+ 4+ 13= (s+2)²+ 13.

And yes, factoring the e^-3s out is a good idea.

Of course you can factor it - you just can't factor it over the reals!

After you do the factoring (yes, with complex factors!) then you can rewrite using partial fractions and look up the inverse transforms in tables. Or you can just use Cauchy's Residue Theorem directly since you already know where the poles are.

 

[Theelectricchild]

I was foolish, it's a very easy problem once you learn the step function. The e^-3s is just a shift, using the u(t) the shift is 3.