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taurus [48]
3 years ago
11

Lila knows that 316 means “3 divided by 16.” She uses this to find the decimal equivalent for 316. Enter a digit into each box t

o continue her work.

Mathematics
1 answer:
jeka57 [31]3 years ago
6 0

He first took six pieces for himself and then evenly divide… Get the answers you need, ... Log in to add comment. swebster1495 is ... Lila knows that 316 means “3 divided by 16.” She uses this to find the decimal equivalent for 316. Enter a digit into each box to continue her work. All the ratios lie on line .

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B-2-11. Find the inverse Laplace transform of s + 1/s(s^2 + s +1)
Aleksandr-060686 [28]

Answer:

\mathcal{L}^{-1}\{\frac{s+1}{s(s^{2} + s +1)}\}=1-e^{-t/2}cos(\frac{\sqrt{3} }{2}t )+\frac{e^{-t/2}}{\sqrt{3} }sin(\frac{\sqrt{3} }{2}t)

Step-by-step explanation:

let's start by separating the fraction into two new smaller fractions

.

First,<em> s(s^2+s+1)</em> must be factorized the most, and it is already. Every factor will become the denominator of a new fraction.

\frac{s+1}{s(s^{2} + s +1)}=\frac{A}{s}+\frac{Bs+C}{s^{2}+s+1}

Where <em>A</em>, <em>B</em> and <em>C</em> are unknown constants. The numerator of <em>s</em> is a constant <em>A</em>, because <em>s</em> is linear, the numerator of <em>s^2+s+1</em> is a linear expression <em>Bs+C</em> because <em>s^2+s+1</em> is a quadratic expression.

Multiply both sides by the complete denominator:

[{s(s^{2} + s +1)]\frac{s+1}{s(s^{2} + s +1)}=[\frac{A}{s}+\frac{Bs+C}{s^{2}+s+1}][{s(s^{2} + s +1)]

Simplify, reorganize and compare every coefficient both sides:

s+1=A(s^2 + s +1)+(Bs+C)(s)\\\\s+1=As^{2}+As+A+Bs^{2}+Cs\\\\0s^{2}+1s^{1}+1s^{0}=(A+B)s^{2}+(A+C)s^{1}+As^{0}\\\\0=A+B\\1=A+C\\1=A

Solving the system, we find <em>A=1</em>, <em>B=-1</em>, <em>C=0</em>. Now:

\frac{s+1}{s(s^{2} + s +1)}=\frac{1}{s}+\frac{-1s+0}{s^{2}+s+1}=\frac{1}{s}-\frac{s}{s^{2}+s+1}

Then, we can solve the inverse Laplace transform with simplified expressions:

\mathcal{L}^{-1}\{\frac{s+1}{s(s^{2} + s +1)}\}=\mathcal{L}^{-1}\{\frac{1}{s}-\frac{s}{s^{2}+s+1}\}=\mathcal{L}^{-1}\{\frac{1}{s}\}-\mathcal{L}^{-1}\{\frac{s}{s^{2}+s+1}\}

The first inverse Laplace transform has the formula:

\mathcal{L}^{-1}\{\frac{A}{s}\}=A\\ \\\mathcal{L}^{-1}\{\frac{1}{s}\}=1\\

For:

\mathcal{L}^{-1}\{-\frac{s}{s^{2}+s+1}\}

We have the formulas:

\mathcal{L}^{-1}\{\frac{s-a}{(s-a)^{2}+b^{2}}\}=e^{at}cos(bt)\\\\\mathcal{L}^{-1}\{\frac{b}{(s-a)^{2}+b^{2}}\}=e^{at}sin(bt)

We have to factorize the denominator:

-\frac{s}{s^{2}+s+1}=-\frac{s+1/2-1/2}{(s+1/2)^{2}+3/4}=-\frac{s+1/2}{(s+1/2)^{2}+3/4}+\frac{1/2}{(s+1/2)^{2}+3/4}

It means that:

\mathcal{L}^{-1}\{-\frac{s}{s^{2}+s+1}\}=\mathcal{L}^{-1}\{-\frac{s+1/2}{(s+1/2)^{2}+3/4}+\frac{1/2}{(s+1/2)^{2}+3/4}\}

\mathcal{L}^{-1}\{-\frac{s+1/2}{(s+1/2)^{2}+3/4}\}+\mathcal{L}^{-1}\{\frac{1/2}{(s+1/2)^{2}+3/4}\}\\\\\mathcal{L}^{-1}\{-\frac{s+1/2}{(s+1/2)^{2}+3/4}\}+\frac{1}{2} \mathcal{L}^{-1}\{\frac{1}{(s+1/2)^{2}+3/4}\}

So <em>a=-1/2</em> and <em>b=(√3)/2</em>. Then:

\mathcal{L}^{-1}\{-\frac{s+1/2}{(s+1/2)^{2}+3/4}\}=e^{-\frac{t}{2}}[cos\frac{\sqrt{3}t }{2}]\\\\\\\frac{1}{2}[\frac{2}{\sqrt{3} } ]\mathcal{L}^{-1}\{\frac{\sqrt{3}/2 }{(s+1/2)^{2}+3/4}\}=\frac{1}{\sqrt{3} } e^{-\frac{t}{2}}[sin\frac{\sqrt{3}t }{2}]

Finally:

\mathcal{L}^{-1}\{\frac{s+1}{s(s^{2} + s +1)}\}=1-e^{-t/2}cos(\frac{\sqrt{3} }{2}t )+\frac{e^{-t/2}}{\sqrt{3} }sin(\frac{\sqrt{3} }{2}t)

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