Question

Find the Laplace transform using the definition y{f(t)} = ∫(0 to [infinity]) e^(-st) f(t) dt.

(1) f(t) = 1 - t
(2) f(t) = 1 - t
(3) f(t) = {1 for t < 0, 2t for t ≥ 0}

Answer 1

The Laplace transform of f(t) = 1 - t is Y{f(t)} = 1/s - 1/s².

The Laplace transform of f(t) = 1 - t is Y{f(t)} = 1/s - 1/s².

The Laplace transform of f(t) = {1 for t < 0, 2t for t ≥ 0} is Y{f(t)} = 1/s + 2/s².

Step-by-step explanation:

For f(t) = 1 - t, applying the Laplace transform definition involves integrating e^(-st)(1 - t) from 0 to ∞. The result is Y{f(t)} = 1/s - 1/s².

The repeated entry f(t) = 1 - t yields the same Laplace transform, Y{f(t)} = 1/s - 1/s². This repetition appears to be a typographical error.

For the piecewise function f(t) = {1 for t < 0, 2t for t ≥ 0}, apply the Laplace transform separately for each interval. The Laplace transform is Y{f(t)} = 1/s + 2/s².

In summary, the Laplace transforms are obtained using the definition Y{f(t)} = ∫(0 to ∞) e^(-st) f(t) dt, resulting in the expressions 1/s - 1/s² for the first two functions and 1/s + 2/s² for the piecewise function.