The Laplace transform of a function f(t) = e^(-at)g(t) where g(t) is any function of t, is given by L{f(t)} = F(s+a) where F(s) is the Laplace transform of g(t).

In this case, we have f(t) = e^(-2t)cos(t). Here, a = 2 and g(t) = cos(t).

The Laplace transform of cos(t) is F(s) = s/(s^2 + 1).

Therefore, the Laplace transform of f(t) = e^(-2t)cos(t) is given by F(s+2) = (s+2)/((s+2)^2 + 1).

So, the Laplace transform of e^(-2t)cos(t) is (s+2)/((s+2)^2 + 1).

Question

The Laplace transform of a function f(t) = e^(-at)g(t) where g(t) is any function of t, is given by L{f(t)} = F(s+a) where F(s) is the Laplace transform of g(t).

In this case, we have f(t) = e^(-2t)cos(t). Here, a = 2 and g(t) = cos(t).

The Laplace transform of cos(t) is F(s) = s/(s^2 + 1).

Therefore, the Laplace transform of f(t) = e^(-2t)cos(t) is given by F(s+2) = (s+2)/((s+2)^2 + 1).

So, the Laplace transform of e^(-2t)cos(t) is (s+2)/((s+2)^2 + 1).

Knowee AI · Accepted Answer

The Laplace transform of a function f(t) = e^(-at)g(t) where g(t) is any function of t, is given by L{f(t)} = F(s+a) where F(s) is the Laplace transform of g(t).

In this case, we have f(t) = e^(-2t)cos(t). Here, a = 2 and g(t) = cos(t).

The Laplace transform of cos(t) is F(s) = s/(s^2 + 1).

Therefore, the Laplace transform of f(t) = e^(-2t)cos(t) is given by F(s+2) = (s+2)/((s+2)^2 + 1).

So, the Laplace transform of e^(-2t)cos(t) is (s+2)/((s+2)^2 + 1).

The Laplace transform of e−2tcos(t)𝑒−2𝑡cos⁡(𝑡) isss+2𝑠𝑠+2s+2s2+1𝑠+2𝑠2+12s2+2s+52𝑠2+2𝑠+5s+2s2+4s+5

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