What is inverse oflaplace (-0.5)/(s^2+s+1) ?

The answer to inverse oflaplace (-0.5)/(s^2+s+1) is -0.57735…e^{-t/2}sin(0.86602…t)

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inverse oflaplace (-0.5)/(s^2+s+1)

inverse laplace

\frac{- 0.5}{s ^{2} + s + 1}

- 0.57735 \dots e^{- \frac{t}{2}} sin (0.86602 \dots t)

Solution steps

L^{- 1} {\frac{- 0.5}{s ^{2} + s + 1}}

Expand

\frac{- 0.5}{s ^{2} + s + 1} : - 0.5 \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{3}{4}}

= L^{- 1} {- 0.5 \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{3}{4}}}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= - 0.5 L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{3}{4}}}

L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{3}{4}}} : e^{- \frac{t}{2}} \frac{2}{3} sin (\frac{3 t}{2})

= - 0.5 e^{- \frac{t}{2}} \frac{2}{3} sin (\frac{3 t}{2})

Simplify

- 0.5 e^{- \frac{t}{2}} \frac{2}{3} sin (\frac{3 t}{2}) : - 0.57735 \dots e^{- \frac{t}{2}} sin (0.86602 \dots t)

= - 0.57735 \dots e^{- \frac{t}{2}} sin (0.86602 \dots t)

\int x - 2 + \frac{3}{( 3 x - 5 ) ^{6}} d x

\frac{t}{t ^{2} + 8}

(2 x - 7) 4 (x^{2} + x + 1) 5

\frac{d}{d x} (- x + 2 cos (2 x))

y = \frac{4 x}{x + 2}, (2, 2)

What is inverse oflaplace (-0.5)/(s^2+s+1) ?
The answer to inverse oflaplace (-0.5)/(s^2+s+1) is -0.57735…e^{-t/2}sin(0.86602…t)