Question

Consider the initial value function y given by

Answer 1

Answer:

y(s) = $\frac{5s-53}{s^{2} - 10s + 26}$

we will compare the denominator to the form $(s-a)^{2} +\beta ^{2}$

$s^{2} -10s+26 = (s-a)^{2} +\beta ^{2} = s^{2} -2as +a^{2} +\beta ^{2}$

comparing coefficients of terms in s

$s^{2} :$ 1

s: -2a = -10

      a = -2/-10

      a = 1/5

constant: $a^{2}+\beta ^{2} = 26$

               $(\frac{1}{5} )^{2} + \beta ^{2} = 26\\\\\beta^{2} = 26 - \frac{1}{10} \\\\\beta =\sqrt{26 - \frac{1}{10}} =5.09$

hence the first answers are:

a = 1/5 = 0.2

β = 5.09

Given that y(s) = $A(s-a)+B((s-a)^{2} +\beta ^{2} )$

we insert the values of a and β

  $\\5s-53$ = $A(s-0.2)+B((s-0.2)^{2} + 5.09^{2} )$

to obtain the constants A and B we equate the numerators and we substituting s = 0.2 on both side to eliminate A

5(0.2)-53 = A(0.2-0.2) + B((0.2-0.2)²+5.09²)

-52 = B(26)

B = -52/26 = -2

to get A lets substitute s=0.4

5(0.4)-53 = A(0.4-0.2) + (-2)((0.4 - 0.2)²+5.09²)

-51 = 0.2A - 52.08

0.2A = -51 + 52.08

A = -1.08/0.2 = 5.4

the constants are

a = 0.2

β = 5.09

A  = 5.4

B = -2

Step-by-step explanation:

1. since the denominator has a complex root we compare with the standard form $s^{2} -10s+26 = (s-a)^{2} +\beta ^{2} = s^{2} -2as +a^{2} +\beta ^{2}$
2. Expand and compare coefficients to obtain the values of a and β as shown above
3. substitute the values gotten into the function
4. Now assume any value for 's' but the assumption should be guided to eliminate an unknown, just as we've use s=0.2 above to eliminate A
5. after obtaining the first constant, substitute the value back into the function and obtain the second just as we've shown clearly above

Thanks...