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Find the Laplace transform of each of the following five functions.(1)flt) = t(4) + 3t3 _ t2 + 2t...

Question

Find the Laplace transform of each of the following five functions.(1)flt) = t(4) + 3t3 _ t2 + 2t

Find the Laplace transform of each of the following five functions. (1) flt) = t(4) + 3t3 _ t2 + 2t



Answers

Determine the Laplace transform of each of the following functions:
(a) $f_{1}(t)=$ $\left\{\begin{array}{ll}{0,} & {t=2,} \\ {t,} & {t \neq 2.}\end{array}\right.$
(b) $f_{2}(t)=$ $\left\{\begin{array}{ll}{5,} & {t=1,} \\ {2,} & {t=6,} \\ {t,} & {t \neq 1,6.}\end{array}\right.$
(c) $f_{3}(t)=t.$
Which of the preceding functions is the inverse Laplace transform of 1$/ s^{2} ?$

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Hi. Today we're discussing the black last transform off three functions. What's common to all the function is that they're equal to each of the tea. Accepts asked financially many points, which means that all into girls of thumb are going to be equal in particular. This means that the lack last transform of any of the three functions given is gonna be equal to the lap. Last transform off U to the T, which is equal to one over X minus one when we're looking at the end of US lack last transformer malfunction. We're actually looking for the continuous function. Who's like last transformers? One of ass minus one. So the actual act last transformer is gonna be to the tea or the third function Give him because it's the only continuous function. Yeah,

Hi. Today we're looking at the last blast transom of three different functions. If we look at the question, though, what's common to all of them is that they are equal to each of the tea. Except for finally many values. T since the last last transformer only depends on the values of the function except of finally many points. We know that the last trip it transforms any of the three functions is gonna be equal to one over X minus one, which is the same as the last blast. Transform off each of the tea. However, the inverse lap last transform of one over X minus one. It's only equal to eat the tea because the end of a slap US transform always takes the continuous function. Who's like last chance? One is the function given and this function e to the T, which is the third function, is the only function which is continuous at the three


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