F (s) is the Laplace transform of the functionf (t) = 2t2e-tF(1) is __...
Laplace Transform and its Definition:
The Laplace transform is a mathematical operation that transforms a function of a real variable t into a function of a complex variable s. It is commonly used in engineering and physics to solve differential equations and analyze systems in the frequency domain.
The Laplace transform of a function f(t), denoted as F(s), is defined as:
F(s) = L{f(t)} = ∫[0 to ∞] e^(-st) f(t) dt
where s is a complex variable and t is the time variable.
Given Function and its Laplace Transform:
In this problem, we are given the function f(t) = 2t^2e^(-t) and we need to find its Laplace transform F(s).
To find the Laplace transform, we substitute the given function into the Laplace transform definition:
F(s) = L{f(t)} = ∫[0 to ∞] e^(-st) (2t^2e^(-t)) dt
Simplifying the expression inside the integral:
F(s) = 2 ∫[0 to ∞] t^2 e^(-st) e^(-t) dt
= 2 ∫[0 to ∞] t^2 e^(-(s+1)t) dt
Integration by Parts:
To solve this integral, we can use the technique of integration by parts. The formula for integration by parts is:
∫u dv = uv - ∫v du
We choose u = t^2 and dv = e^(-(s+1)t) dt. Then, we can find du and v by differentiating and integrating u and dv, respectively.
du = 2t dt
v = -1/(s+1) e^(-(s+1)t)
Substituting these values into the integration by parts formula:
∫ t^2 e^(-(s+1)t) dt = (-t^2/(s+1)) e^(-(s+1)t) - ∫ (-2t/(s+1)) e^(-(s+1)t) dt
Recursive Integration by Parts:
We can see that we have a similar integral on the right-hand side of the equation. We can apply integration by parts again to solve it.
Let's denote the second integral as I. We choose u = -2t/(s+1) and dv = e^(-(s+1)t) dt. Then, we can find du and v by differentiating and integrating u and dv, respectively.
du = -2/(s+1) dt
v = -1/(s+1) e^(-(s+1)t)
Substituting these values into the integration by parts formula:
I = (-2t/(s+1)) e^(-(s+1)t) - ∫ (-2/(s+1)) (-1/(s+1)) e^(-(s+1)t) dt
= (-2t/(s+1)) e^(-(s+1)t) + (2/(s+1))^2 e^(-(s+1)t) - ∫ (2/(s+1))^2 e^(-(s+1)t) dt
We can observe that we have another similar integral on the right-hand side.
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