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3 years ago

Determine whether the expressions are equivalent to

Mathematics

1 answer:

Leviafan [203]3 years ago

6 0

Answer:

Not equivalent

Step-by-step explanation:

Because that's way more

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Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de

Ganezh [65]

Answer:

a. $\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

b. $\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

Step-by-step explanation:

The initial value problem is given as:

$y' +y = 7+\delta (t-3) \\ \\ y(0)=0$

Applying laplace transformation on the expression $y' +y = 7+\delta (t-3)$

to get $L[{y+y'} ]= L[{7 + \delta (t-3)}]$

$l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

Taking inverse of Laplace transformation

$y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$

$L^{-1}[\dfrac{e^{-3s}}{s+1}]$

$L^{-1}[\dfrac{1}{s+1}] = e^{-t} = f(t) \ then \ by \ second \ shifting \ theorem;$

$L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t$

$L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t$

$= e^{-t-3} \left \{ {{1 \ \ \ \ \ t>3} \atop {0 \ \ \ \ \ t$

= $e^{-(t-3)} u (t-3)$

Recall that:

$y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$

Then

$y(t) = 7 -7e^{-t} +e^{-(t-3)} u (t-3)$

$y(t) = 7 -7e^{-t} +e^{-t} e^{-3} u (t-3)$

$\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

3 0

3 years ago

A bank loaned out $19,500, part of it at the rate of 11% annual interest, and the rest at 7% annual interest. The total interest

Morgarella [4.7K]

The amount loaned at the interest rate of 11% is $3000.

The amount loaned at the interest rate of 7% is $16,500

<h3>What are the linear equations that represent the question?
a + b = 19500 equation 1</h3>

0.11a + 0.07b = 1485 equation 2

Where:

a = amount loaned at the interest rate of 11%

b = = amount loaned at the interest rate of 7%

<h3>How much was loaned at 11% and 7%</h3>

In order to determine how much was loaned at 7%, take the following steps:

Multiply equation 1 by 0.11:

.11a + 0.11b = 2145 equation 3

Subtract equation 2 from equation 3:

0.04b = 660

Divide both sides by 0.04:

b = 16,500

Subtract 16,500 from 19,500 :

19,500 - 16,500 = $3000

To learn more about interest, please check: brainly.com/question/26164549

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7 0

2 years ago

The walls of a square room need painting. each of the four walls is 12ft wide by 8.5 feet tall. one gallon of paint covers 75fts

miv72 [106K]

The surface to be paint is 4 * 12*8.5= 408 ft squared

1 gallon.......75 ftsquared
x gallon.......408 ft squared
------------------------------------
x= 408/75= 5,44 gallons

7 0

3 years ago

A natural number that is not a rational number

zimovet [89]

3.1456493394949

4.29484728283
The number are repeated forever is not rational number

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3 years ago

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Who should fill out the W-2 form?

elena-14-01-66 [18.8K]

It’s A THE ANSWER IS A

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3 years ago

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