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

The answer to inverse oflaplace (s^2+s)/(s^2+s+1) is \delta(t)-e^{-t/2} 2/(sqrt(3))sin((sqrt(3)t)/2)

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

L^{- 1} {1} : δ (t)

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

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

y^{^{''}} - 3 y^{^{'}} + 7 y = x e^{x}

\int \frac{8}{12 - x ^{2} - 4 x} d x

3 x^{2}

\int (2 x + 1)^{7} d x

\frac{d}{d x} (ln^{2} (sin (2 x)))

What is inverse oflaplace (s^2+s)/(s^2+s+1) ?
The answer to inverse oflaplace (s^2+s)/(s^2+s+1) is \delta(t)-e^{-t/2} 2/(sqrt(3))sin((sqrt(3)t)/2)