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# Laplace inverse of (s^2+2)/(s^4+s^2+1)?

What is the Inverse Laplace transforms of (s^2+2)/(s^4+s^2+1)

### 1 Answer

- kbLv 76 years agoFavorite Answer
Note that

s^4 + s^2 + 1

= (s^4 + 2s^2 + 1) - s^2

= (s^2 + 1)^2 - s^2

= (s^2 + 1 + s)(s^2 + 1 - s), via difference of squares.

So, partial fractions yields

(s^2 + 2)/(s^4 + s^2 + 1)

= (As + B)/(s^2 + s + 1) + (Cs + D)/(s^2 - s + 1)

Clearing denominators to solve for the constants,

s^2 + 2 = (As + B)(s^2 - s + 1) + (Cs + D)(s^2 + s + 1)

............= (A+C)s^3 + (-A+B+C+D)s^2 + (A-B+C+D)s + (B+D)

Equate like coefficients:

A + C = 0

-A + B + C + D = 1

A - B + C + D = 0 ==> -B + D = 0 (from the first equation)

B + D = 2.

Solving yields B = D = 1 and A = 1/2, C = -1/2

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So, (s²+2)/(s⁴+s²+1)

= (1/2) [(s+2)/(s²+s+1) + (-s+2)/(s²-s+1)]

= (1/2) [(s+2)/((s + 1/2)² + 3/4) + (-s+2)/((s - 1/2)² + 3/4)]

= (1/2) [((s + 1/2) + 3/2)/((s + 1/2)² + 3/4) + (-(s - 1/2) + 3/2)/((s - 1/2)² + 3/4)]

Therefore,

L⁻¹{(s²+2)/(s⁴+s²+1)}

= e^(-t/2) * L⁻¹{(1/2) (s + 3/2)/(s² + 3/4)}

+ e^(t/2) L⁻¹{(1/2) (-s + 3/2)/(s² + 3/4)}, via time shifting theorem

= (1/2)e^(-t/2) [cos(t√3/2) + (3/2)(2/√3) sin(t√3/2)]

+ (1/2) e^(t/2) [-cos(t√3/2) + (3/2)(2/√3) sin(t√3/2)]

= (1/2) e^(-t/2) [cos(t√3/2) + √3 sin(t√3/2)]

+ (1/2) e^(t/2) [-cos(t√3/2) + √3 sin(t√3/2)]

= (-1/2)(e^(t/2) - e^(-t/2)) cos(t√3/2) + (1/2)(e^(t/2) + e^(-t/2)) * √3 sin(t√3/2)]

= -sinh(t/2) cos(t√3/2) + √3 cosh(t/2) sin(t√3/2).

(Double checked on wolfram alpha.)

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I hope this helps!