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Laplace transfrom of e^1/2 sint ×cos4t



User Irezwi
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Answer: To find the Laplace transform of e^(1/2) sin(t) * cos(4t), we use the properties of the Laplace transform and the formula for the Laplace transform of sin(at) and cos(bt).

Using the formula for the Laplace transform of sin(at):

L{sin(at)} = a / (s^2 + a^2)

And the formula for the Laplace transform of cos(bt):

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

We can write e^(1/2) sin(t) * cos(4t) as:

e^(1/2) * sin(t) * cos(4t) = (1/2) * (e^(1/2) * sin(5t) - e^(1/2) * sin(3t))

To find the Laplace transform of each term, we can use linearity property of the Laplace transform:

L{e^(1/2) * sin(t) * cos(4t)} = (1/2) * (L{e^(1/2) * sin(5t)} - L{e^(1/2) * sin(3t)})

Applying the Laplace transform to each term:

L{e^(1/2) * sin(5t)} = 1 / (s^2 + (5/2)^2)

L{e^(1/2) * sin(3t)} = 1 / (s^2 + (3/2)^2)

So the Laplace transform of e^(1/2) sin(t) * cos(4t) is:

L{e^(1/2) * sin(t) * cos(4t)} = (1/2) * (1 / (s^2 + (5/2)^2) - 1 / (s^2 + (3/2)^2))

User Flytofuture
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