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first factor s^2+5s+6 and post what you got

if H(s) = 1/s^2+5s+6
and F(s)=1/(s+2)
and G(s)=1/(s+3), then H(s) = F(s)G(s)

so, h(t) = f(t)*g(t) = \[\int\limits_{0}^{t}f(\tau)g(t-\tau) d \tau\]

if F(s) = 1/(s+2), then inverse laplace transform of it is f(t) = e^(-2t)

you have all the information you need to solve this.

so final answer after integrating would be 3^(-2t)-e^(-3t)

I ment to say e^(-2t)-e^-3t

http://www.youtube.com/watch?v=TJgBEI3drUc&feature=BFa&list=SP96AE8D9C68FEB902&index=44

thank you just checked inm solution and it is

okay great!

\[H(s)=1/s(s^2+9)=A/s+Bs+C/(s^2+9) by \partial fraction I get (1/9s) -(s/9(s^2+9)\]

you can't express them as a summation.

\[so \Im \left with 1/9 L^-1 (1/s) -1/9 L^-1(s/s^2+9)\]

so F(s)=1/9 and G(s)=-1/9cos3t

\[\int\limits_{0}^{t}1/9-1/9\cos3rhod \rho= ?\]

I got answer 1/3sin3t and book answer is (1-cos3t)/9

cool