Another Laplace Transform Question

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SUMMARY

The discussion focuses on finding the Laplace transform of the function f(t) = -9t + 7 + u5(-7 + 9t + et). The Laplace transforms of the first two terms are calculated as -9/s² and 7/s, respectively. For the term involving the unit step function u5, the transformation is handled using the formula L[u_c(f(t))] = e^{-cs}L[f(t+c)], resulting in the final expression: -9/s² + 7/s + e^{5s}(-2/s + 9/s² + 1/s). It is emphasized that the Laplace transform is a function of s, while the inverse is a function of t.

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Homework Statement


Find the Laplace transform of f(t).

f(t) = -9t + 7 + u5(-7 + 9t + et)


Homework Equations





The Attempt at a Solution


The Laplace transform of -9t is just -9/t2

The Laplace transform of 7 is 7/s

But I am not sure how to find it for u5(-7 + 9t + et). In the tables it says ua(t)f(t-a) = e-asF(s). Is that the line I should use? Even if it is, I don't know how.
 
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You've got the right start. You want to take it term by term, and you have the Laplace transform down for the first two terms. The way to handle step functions is as follows. First, get rid of the step function by saying:

[tex]L[u_c(f(t))] = e^{-cs}L[f(t+c)][/tex]

So in your case, the third term would become

[tex]e^{-5s} L[-2 + 9t + e^t][/tex]

Now it's just another term by term Laplace transform. So your final answer should be

[tex] -9/s^2 + 7/s + e^{5s}(-2/s + 9/s^2 + 1/s)[/tex]

n.b.: Remember that the Laplace transform is always a function of s, while the inverse Laplace transform is a function of t. I see you got it slightly mixed up, so just something to keep in mind! :)
 

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