Consider the initial value function y given by

y"-10y'+26y=0, y(0) = 5, y'(0) = -3, Y(s) = (5s-53)/(s^2-10s+26)The characteristic polynomial in the denominator of Y(s) above has complex roots. Therefore, complete the square in that denominator and rewrite Y(s) as follows, Y(s)=A(s−α)+B((s−α)2+β2) where A, B,α,β are constants, with β⩾0. Find the the constants A, B,α,β.

y(s) =

we will compare the denominator to the form

comparing coefficients of terms in s

1

s: -2a = -10

      a = -2/-10

      a = 1/5

constant:

hence the first answers are:

a = 1/5 = 0.2

β = 5.09

Given that y(s) =

we insert the values of a and β

  =

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:

Thanks...

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