Inverse laplace transform (easy one)

[Derill03]

I am asked to find inverse laplace of (3s+5)/(s^2-6s+25)

I do partial frac and get (3s)/((s-3)^2+16) + (5/(s-3)^2+16)

I do the translation and get [(3s)/s^2+16 + 5/(s^2+16)]*e^3t

My final answer is e^3t[3cos(4t) + 5/4sin(4t)]

but the book says the sin term should be 7/2sin(4t)? I say 5/4 because a 4 needs to be in numerator to make use of the table for k/(s^2+k^2) so if you multiply 5 by (5/4) the 4 can be in numerator as 4/(s^2+16) because it will cancel and be the original 5.

I just need to know how the 7/2 gets in the sin term and why its not (5/4)?

 

[vela]

You didn't handle the shift correctly. The bottom has (s-3) in it, but the top doesn't for the cosine term. Fixing that will probably fix the sine term.