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Sagot :
The Laplace transform of the given function is [tex]f(s)=\frac{1-e^{-5s}(1-5s) }{s^{2} }[/tex]
Given,
[tex]f(t)=\left \{ {{t, 0\leq t < 5} \atop {0,t\geq 5}} \right.[/tex]
Write the function in unit step function.
[tex]f(t)=t(u_{0}(t)-u_{5} (t))[/tex]
= [tex]t(1-u_{5}(t))[/tex] ∵[tex]u_{0} (t)=1[/tex]
= [tex]t-tu_{5} (t)[/tex]
[tex]f(t)=t-tu_{5} (t)[/tex]
In order to make it easier to take the Laplace transform of the function, we can follow the steps:
[tex]f(t)=t-tu_{5} (t)[/tex]
= [tex]t-(t-5+5)u_{5} (t)[/tex]
= [tex]t-(t-5)u_{5} (t)+5u_{5} (t)[/tex]
[tex]f(t)=t-(t-5)u_{5} (t)+5u_{5} (t)[/tex]
We need to find the Laplace transform of f(t).
Apply Laplace transform on both sides,
£[tex][f(t)=[/tex]£[tex](t)-[/tex]£[tex](t-5)u_{5} (t))+5[/tex]£[tex](u_{5} (t))[/tex]
= [tex]\frac{1}{s^{2} } -e^{-s}[/tex]£[tex](t)+5(\frac{e^{-5s} }{s} )[/tex]
= [tex]\frac{1}{s^{2} } -e^{-5s} (\frac{1}{s^{2} } )+\frac{5se^{-5s} }{s^{2} }[/tex]
= [tex]\frac{1-e^{-5s}+5se^{-5s} }{s^{2} }[/tex]
= [tex]\frac{1-e^{-5s}(1-5s) }{s^{2} }[/tex]
[tex]f(s) =\frac{1-e^{-5s}(1-5s) }{s^{2} }[/tex]
Learn more about Laplace transform here: https://brainly.com/question/17190535
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