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t( e^[-2t])u(t)
where u(t) is the unit step function
u(t) =0 for t<0 seconds
and 1 when this equal to or greater than 0 seconds
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to find the inverse transform of
1/(s +2)^2
do a partial fraction expansion if possible
1/(s +2)^2 = A/(s +2) + B/( s+ 2)^2
1= As +2A + B
A = 0
and B =1
From the above
1/(s +1)^2 is as simplified as it can be
Consult a table of Laplace transforms
The inverse transform of 1/( s +a)^2 is
t(e^[-at])(u(t)
where u(t) is the unit step function
u(t) = 0 when t< 0 seconds
and 1 when t is dual to or greater than 0 seconds
Step-by-step explanation:
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