Question 1185331: Find the inverse Laplace transform f(t)=L^(−1){F(s)} of the function
F(s)=(e^(−s)(6s−5))/(s^2+64)
You may use h(t) for the Heaviside step function.
f(t)=L^(−1){(e^(−s)(6s−5))/(s^2+64)}=
My answer is h(t-1)(6cos(8h(t-1)))-5/8sin(8h(t-1)) , wrong
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to find the inverse Laplace transform of F(s) = (e^(-s)(6s - 5))/(s² + 64):
1. **Recognize the time shift:** The term e^(-s) indicates a time shift. Recall that L{f(t-a)u(t-a)} = e^(-as)F(s), where u(t-a) is the Heaviside step function. In our case, a = 1. This means we'll have a function shifted by 1 unit in time.
2. **Inverse transform of the rest:** We need to find the inverse Laplace transform of G(s) = (6s - 5)/(s² + 64). We can break this into two parts:
* L⁻¹{6s/(s² + 64)} = 6L⁻¹{s/(s² + 8²)} = 6cos(8t)
* L⁻¹{-5/(s² + 64)} = -5L⁻¹{1/(s² + 8²)} = (-5/8)L⁻¹{8/(s² + 8²)} = (-5/8)sin(8t)
Therefore, L⁻¹{G(s)} = 6cos(8t) - (5/8)sin(8t)
3. **Apply the time shift:** Since F(s) = e^(-s)G(s), we have f(t) = g(t-1)u(t-1), where g(t) = 6cos(8t) - (5/8)sin(8t).
So, f(t) = [6cos(8(t-1)) - (5/8)sin(8(t-1))]u(t-1)
Therefore, the inverse Laplace transform is:
f(t) = [6cos(8(t-1)) - (5/8)sin(8(t-1))]u(t-1)
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