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Find the inverse Laplace transform?

For e^(-10s)/(s(s^2+3s+2)).

The answer is u10(t)[(1/2)+(1/2)e^(-2(t-10))-e^(-(t-10))]

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George C. | Humboldt State and Georgetown graduateHumboldt State and Georgetown graduate
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e^(-10s)/(s(s^2+3s+2))

Factor out e^(-10s) for partial fraction decomposition.

1/(s(s^2+3s+2))  =   1/ s(s+1)(s+2)  =   A/s  +  B/(s+1)  +  C/(s+2)

Using the Heaviside coverup method.

when s = 0,   A= 1/ (  )(0 + 1)(0 + 2)  =  1/2

when s = -1, B= 1/(-1)(  )(-1 + 2)  =  -1

When s = -2, C= 1/(-2)(-2+1)(  )   =  1/2

  So  (1/2)/s  +   (-1)/(s + 1)   +   (1/2)/(s + 2)

=  (1/2)/(s - 0)   +   (-1)/(s + 1)   +   (1/2)/(s + 2)

=  (1/2) e^(0*t)   -  e^(-t)    +   (1/2) e^(-2t)

=  (1/2                -  e^-(t)     +   (1/2)  e^-2(t))

And u(c)(t)*f(t - c)  =  e^(-cs)*F(s) = e^(-10s)*F(s)  =  u(10)(t)*f(t - 10)

= u(10)(t)[ (1/2)   -   e^-(t - 10)    +    (1/2) e^ -2(t  -  10)]