Determine the inverse laplace transforms of
s+7/s^2+4s+3
and
3s-8/s^2+4
1: Complete the square down below and you can split it into two LTs of the form:
multiplied by whatever constants. Or they could be - a^2 on the bottom. Whatever, you use your inverse tables and get or it could be cos or sinh or cosh, whatever, see what you get when you rip it apart.
You can simply factor:
1. (s+7)/[s^2+4s+3] = (s+7)/[(s+3)(s+1)]
= (s+3)/[[(s+3)(s+1)] + 4/[(s+3)(s+1)]
= 1/(s+1) + 4/[(s+3)(s+1)]
y(t) = e^(-t ) + 4 (e^(-3t)-e^(-t))/-2
y(t) = 3e^(-t) - 2e^(-3t)
2. write as 3s/(s^2+4) - 8 /(s^2+4) = 3{s/(s^2+4)} - 4{2 /(s^2+4)}
y(t) = 3cos(2t) - 4sin(2t)