use properties of the Laplace transform and the table of Laplace transforms to determine L[f] f(t)=2(t-5)u_5(t)

Efan Halliday 2021-01-13 Answered

use properties of the Laplace transform and the table of Laplace transforms to determine L[f]

f (t) = 2 (t - 5) u_{5} (t)

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Expert Answer

avortarF

Answered 2021-01-14 Author has 113 answers

Step 1

Given:

f (t) = 2 (t - 5) u_{5} (t)

L [f (t)] - ?

Step 2

Formula : L [f (t - c) u_{c} (t)] = e^{- c s} L [f (t)]

Step 3

Solution : f (t) = 2 (t - 5) u_{5} (t)

Take Laplace Transform to both sides:

L [f (t)] = L [2 (t - 5) u_{5} (t)]

L [f (t)] = 2 L [(t - 5) u_{5} (t)]

L [f (t)] = 2 e^{- 5 s} L (t) \Rightarrow [L [f (t - c) u_{c} (t)] = e^{- c s} L [f (t)]]

L [f (t)] = 2 e^{- 5 s} (\frac{1!}{s^{1 + 1}}) \Rightarrow [L (t^{n}) = \frac{n!}{s^{n + 1}}]

L [f (t)] = 2 e^{- 5 s} (\frac{1}{s^{2}})

\begin{array}{cc} L [f (t)] = 2 \frac{e^{- 5 s}}{s^{2}} \end{array}

Step 4

Answer : L [f (t)] = 2 \frac{e^{- 5 s}}{s^{2}}

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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})

We need to analyze where is positive and negative. So I solved the inequalities:

\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)

.
The case of the product being negative implies:

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 .

So the function is increasing in

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

. So the function does not have maximum or minimum. Not sure of this but what do you think?

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.
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.
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f (x) = \frac{1}{(x - 2) (2 x - 5)}

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, we define f(g(x)) over the domain in which its defined , is it correct to say f(g(x)) is discontinuous at

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1 / \sqrt 2

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I am given the function

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and told that routine computations will show that

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x (1) = π

.
I must determine a differential equation for

x (t)

x^{'} (t) = f (t, x)

.
I must then use

f

π

.
I will use Euler's Method to find minimum number of iterations (

n

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π \approx 3.1400

n - 1

iterations is strictly less than 3.1400.
How do I find

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Analysis