Determine the inverse laplace transforms of

s+7/s^2+4s+3

and

3s-8/s^2+4

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- Aug 11th 2009, 12:08 PMtone999Inverse Laplace Transforms
Determine the inverse laplace transforms of

s+7/s^2+4s+3

and

3s-8/s^2+4 - Aug 11th 2009, 12:56 PMMatt Westwood
1: Complete the square down below and you can split it into two LTs of the form:

$\displaystyle \frac {s-b}{(s-b)^2 + a^2}, \frac {1}{(s-b)^2 + a^2}$

multiplied by whatever constants. Or they could be - a^2 on the bottom. Whatever, you use your inverse tables and get $\displaystyle e^{bt} \sin a t / a$ or it could be cos or sinh or cosh, whatever, see what you get when you rip it apart. - Aug 11th 2009, 01:28 PMCalculus26
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)