Question

1) find the loplace transform e-t cos2t
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Answer 1

Step-by-step explanation:

You can apply the definition L{f(t)} = Integral (from 0 to infinity) of e^(-st)f(t)dt

or apply the form on the Laplace Transform for e^(at)cos(bt):

L{e^(at)cos(bt)} = (s-a)/[(s-a)^2+b^2]

The above form could be found by applying a translation by a to the Laplace Transform of cos(bt):

L{e^(at)f(t)} = F(s-a)

since

Integral (from 0 to infinity) e^(at)e^(-st)f(t)dt

= Integral (from 0 to infinity) e^[(s-a)t]f(t)dt

= F(s-a) for a > s

L{cos(bt)} = Integral (0 to infinity) e^(-st)cos(bt)dt

= -1/s{[e^(-st)cos(bt) | (0 to infinity) + b Integral (0 to infinity) e^(-st)sin(bt)dt}

= -1/s{-1 + b (-1/s)[e^(-st)sin(bt) | (0 to infinity) - b Integral (0 to infinity) e^(-st)cos(bt)dt]

= 1/s - (b/s)^2 Integral (0 to infinity) e^(-st)cos(bt)dt

=> Integral (0 to infinity) e^(-st)cos(bt)dt = (1/s)/[1+(b/s)^2]

= (1/s)/[(s^2+b^2)/s^2] = s/(s^2+b^2)

So L{cos(bt)} = s/(s^2+b^2) = F(s)

L{e^(at)cos(bt)} = F(s-a) = (s-a)/[(s-a)^2+b^2]