find the inverse Laplace transform of the given function F(s)=frac{e^{-2}+e^{-2s}-e^{-3s}-e^{-4s}}{s}

Trent Carpenter 2020-10-28 Answered

find the inverse Laplace transform of the given function

F (s) = \frac{e^{- 2} + e^{- 2 s} - e^{- 3 s} - e^{- 4 s}}{s}

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Expert Answer

un4t5o4v

Answered 2020-10-29 Author has 105 answers

Step 1

We have to find the inverse Laplace transform of the given function:

F (s) = \frac{e^{- 2} + e^{- 2 s} - e^{- 3 s} - e^{- 4 s}}{s}

Step 2

L^{- 1} {F} = L^{- 1} {\frac{e^{- s}}{s}} + L^{- 1} {\frac{e^{- 2 s}}{s}} - L^{- 1} {\frac{e^{- 3 s}}{s}} - L^{- 1} {\frac{e^{- 4 s}}{s}} = u (t - 1) + u (t - 2) - u (t - 3) - u (t - 4)

Step 3

Hence, the final answer is:

u (t - 1) + u (t - 2) + u (t - 3) + u (t - 4)

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New questions

If h has positive derivative and

φ

is continuous and positive. Where is increasing and decreasing f
The problem goes specifically like this:
If h is differentiable and has positive derivative that pass through (0,0), and

φ

is continuous and positive. If:

f (x) = h (\int_{0}^{\frac{x^{4}}{4} - \frac{x^{2}}{2}} φ (t) d t) .

Find the intervals where f is decreasing and increasing, maxima and minima.
My try was this:
The derivative of f is given by the chain rule:

f^{'} (x) = h^{'} (\int_{0}^{\frac{x^{4}}{4} - \frac{x^{2}}{2}} φ (t) d t) φ (\frac{x^{4}}{4} - \frac{x^{2}}{2})

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\frac{x^{4}}{4} - \frac{x^{2}}{2} > 0 \land \frac{x^{4}}{4} - \frac{x^{2}}{2} < 0

(- \infty, - \sqrt 2) \cup (\sqrt 2, \infty)

for the first case and

(- \sqrt 2, \sqrt 2)

for the second one. Then (not sure of this part)

h^{'} (\int_{0}^{\frac{x^{4}}{4} - \frac{x^{2}}{2}} φ (t) d t) > 0

φ (\frac{x^{4}}{4} - \frac{x^{2}}{2}) > 0

x \in (- \infty, - \sqrt 2) \cup (\sqrt 2, \infty) .

Also if both h′ and

φ

are negative the product is positive, that's for

x \in (- \sqrt 2, \sqrt 2)

.
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x \in [(- \infty, - \sqrt 2) \cup (\sqrt 2, \infty)] \cap (- \sqrt 2, \sqrt 2) = [(- \infty, - \sqrt 2) \cap (- \sqrt 2, \sqrt 2)] \cup [(\sqrt 2, \infty) \cap (- \sqrt 2, \sqrt 2)] = \emptyset .

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(- \infty, - \sqrt 2), (- \sqrt 2, \sqrt 2), (\sqrt 2, \infty)

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.
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π

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