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larisa [96]
3 years ago
10

PLEASE HELP !! ILL GIVE BRAINLIEST !!

Mathematics
2 answers:
RSB [31]3 years ago
7 0
Idfk who to do it. I can’t even find out how to do my homework
Vinvika [58]3 years ago
5 0

Answer:

<em><u>1</u></em><em><u>)</u></em><em><u>I</u></em><em><u>T</u></em><em><u> </u></em><em><u>I</u></em><em><u>S</u></em><em><u> </u></em><em><u>C</u></em><em><u>O</u></em><em><u>N</u></em><em><u>G</u></em><em><u>U</u></em><em><u>E</u></em><em><u>R</u></em><em><u>E</u></em><em><u>N</u></em><em><u>T</u></em>

<em><u>2</u></em><em><u>)</u></em><em><u> </u></em><em><u>A</u></em><em><u>L</u></em><em><u>T</u></em><em><u>E</u></em><em><u>R</u></em><em><u>N</u></em><em><u>A</u></em><em><u>T</u></em><em><u>I</u></em><em><u>V</u></em><em><u>E</u></em><em><u> </u></em><em><u>I</u></em><em><u>N</u></em><em><u>T</u></em><em><u>E</u></em><em><u>R</u></em><em><u>I</u></em><em><u>O</u></em><em><u>R</u></em><em><u> </u></em><em><u>A</u></em><em><u>N</u></em><em><u>G</u></em><em><u>L</u></em><em><u>E</u></em><em><u>S</u></em><em><u> </u></em><em><u>A</u></em><em><u>R</u></em><em><u>E</u></em><em><u> </u></em><em><u>E</u></em><em><u>Q</u></em><em><u>U</u></em><em><u>A</u></em><em><u>L</u></em><em><u>.</u></em>

<em><u>3</u></em><em><u>)</u></em><em><u>I</u></em><em><u>T</u></em><em><u> </u></em><em><u>C</u></em><em><u>A</u></em><em><u>N</u></em><em><u> </u></em><em><u>B</u></em><em><u>E</u></em><em><u> </u></em><em><u>W</u></em><em><u>R</u></em><em><u>I</u></em><em><u>T</u></em><em><u>T</u></em><em><u>E</u></em><em><u>N</u></em><em><u> </u></em><em><u>A</u></em><em><u>S</u></em><em><u>=</u></em><em><u>=</u></em><em><u>></u></em><em><u>1</u></em><em><u>2</u></em><em><u>X</u></em><em><u>+</u></em><em><u>1</u></em><em><u>5</u></em><em><u>=</u></em><em><u>1</u></em><em><u>5</u></em><em><u>X</u></em>

<em><u>4</u></em><em><u>)</u></em><em><u>1</u></em><em><u>5</u></em><em><u>X</u></em><em><u>-</u></em><em><u>1</u></em><em><u>2</u></em><em><u>X</u></em><em><u>=</u></em><em><u>1</u></em><em><u>5</u></em>

<em><u>3</u></em><em><u>X</u></em><em><u>=</u></em><em><u>1</u></em><em><u>5</u></em>

<em><u>X</u></em><em><u>=</u></em><em><u>5</u></em><em><u>.</u></em>

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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
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