Answer :
The inverse Laplace transform of F(s) = (s² - 2s + 175) / s² is f(t) = 1 - 2/s + 87.5t
To find the inverse Laplace transform of F(s) = (s² - 2s + 175) / s², you can break down the fraction into simpler terms:
F(s) = (s² / s²) - (2s / s²) + (175 / s²)
Now, apply the linearity property of the inverse Laplace transform:
L⁻¹{(s² / s²)} - L⁻¹{(2s / s²)} + L⁻¹{(175 / s²)}
Now, find the inverse Laplace transforms of each term:
L⁻¹{(s² / s²)}:
This is simply 1 because the Laplace transform of 1 is 1/s.
L⁻¹{(2s / s²)}:
First, factor out the constant:
2 * L⁻¹{(s / s²)}
Now, recall that the Laplace transform of tⁿ is (n!) / sⁿ⁺¹, so in this case:
L⁻¹{(s / s²)} = 1/s (by letting n = 1)
Now, multiply by the constant 2:
2 * (1/s) = 2/s
L⁻¹{(175 / s²)}:
Similarly, factor out the constant:
175 * L⁻¹{(1 / s²)}
The Laplace transform of tⁿ is (n!) / sⁿ⁺¹, so in this case:
L⁻¹{(1 / s²)} = 1/2 * t
Now, multiply by the constant 175:
175 * (1/2 * t) = 87.5t
Now, put all the terms back together:
f(t) = 1 - 2/s + 87.5t
So, the inverse Laplace transform of F(s) is:
f(t) = 1 - 2/s + 87.5t
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