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slamgirl [31]
3 years ago
14

The table shows the input and output of a function Table(input:3,4,6,11. Output:5,7,11,21) write an equation, in function nation

, that would fit in the inputs and outputs in the table. Show your work that proves your equation is true for the given data. PLEASE HELP ITS DUE TODAY!!!l
Mathematics
1 answer:
Anni [7]3 years ago
6 0

f(x) = 2x - 1

To obtain a function we require values of x for x = 3, 4, 5 , 6, .....

The output appears to increase in steps of 2

3 → 5

4 → 7

5 → 9

6 → 11

Looking at the consecutive values of y we should see that these are twice the corresponding value of x subtract 1

f(x) = 2x - 1 ← is a possible function

checking the values for x in the input


f(3) =(2 × 3 ) - 1 = 5

f(4) = (2 × 4 ) - 1 = 7

f(6) = (2 × 6 ) - 1 = 11

f(11) = (2 × 11 ) - 1 = 21




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(10 points) Consider the initial value problem y′+3y=9t,y(0)=7. Take the Laplace transform of both sides of the given differenti
Rashid [163]

Answer:

The solution

Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3 t}

Step-by-step explanation:

<u><em>Explanation</em></u>:-

Consider the initial value problem y′+3 y=9 t,y(0)=7

<em>Step(i)</em>:-

Given differential problem

                           y′+3 y=9 t

<em>Take the Laplace transform of both sides of the differential equation</em>

                L( y′+3 y) = L(9 t)

 <em>Using Formula Transform of derivatives</em>

<em>                 L(y¹(t)) = s y⁻(s)-y(0)</em>

  <em>  By using Laplace transform formula</em>

<em>               </em>L(t) = \frac{1}{S^{2} }<em> </em>

<em>Step(ii):-</em>

Given

             L( y′(t)) + 3 L (y(t)) = 9 L( t)

            s y^{-} (s) - y(0) +  3y^{-}(s) = \frac{9}{s^{2} }

            s y^{-} (s) - 7 +  3y^{-}(s) = \frac{9}{s^{2} }

Taking common y⁻(s) and simplification, we get

             ( s +  3)y^{-}(s) = \frac{9}{s^{2} }+7

             y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}

<em>Step(iii</em>):-

<em>By using partial fractions , we get</em>

\frac{9}{s^{2} (s+3} = \frac{A}{s} + \frac{B}{s^{2} } + \frac{C}{s+3}

  \frac{9}{s^{2} (s+3} =  \frac{As(s+3)+B(s+3)+Cs^{2} }{s^{2} (s+3)}

 On simplification we get

  9 = A s(s+3) +B(s+3) +C(s²) ...(i)

 Put s =0 in equation(i)

   9 = B(0+3)

 <em>  B = 9/3 = 3</em>

  Put s = -3 in equation(i)

  9 = C(-3)²

  <em>C = 1</em>

 Given Equation  9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Comparing 'S²' coefficient on both sides, we get

  9 = A s²+3 A s +B(s)+3 B +C(s²)

 <em> 0 = A + C</em>

<em>put C=1 , becomes A = -1</em>

\frac{9}{s^{2} (s+3} = \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}

<u><em>Step(iv):-</em></u>

y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}

y^{-}(s)  =9( \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}) + \frac{7}{s+3}

Applying inverse Laplace transform on both sides

L^{-1} (y^{-}(s) ) =L^{-1} (9( \frac{-1}{s}) + L^{-1} (\frac{3}{s^{2} }) + L^{-1} (\frac{1}{s+3}) )+ L^{-1} (\frac{7}{s+3})

<em>By using inverse Laplace transform</em>

<em></em>L^{-1} (\frac{1}{s} ) =1<em></em>

L^{-1} (\frac{1}{s^{2} } ) = \frac{t}{1!}

L^{-1} (\frac{1}{s+a} ) =e^{-at}

<u><em>Final answer</em></u>:-

<em>Now the solution , we get</em>

Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3t}

           

           

5 0
3 years ago
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