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

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

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

inverse laplace

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

\frac{e ^{t} + 2 e ^{- \frac{t}{2}} cos ( \frac{3 t}{2} )}{3}

Solution steps

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

For

\frac{s}{s ^{2} - \frac{1}{s}} : F (s) = \frac{1}{s ^{2} - \frac{1}{s}}

= \frac{d}{d t} (L^{- 1} {\frac{1}{s ^{2} - \frac{1}{s}}}) + L^{- 1} {\frac{1}{s ^{2} - \frac{1}{s}}} (0)

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

\frac{d}{d t} \frac{1}{3} e^{t} - \frac{1}{3} e^{- \frac{t}{2}} cos (\frac{3 t}{2}) + \frac{e ^{- \frac{t}{2}} sin ( \frac{3 t}{2} )}{3} = \frac{e ^{t} + 2 e ^{- \frac{t}{2}} cos ( \frac{3 t}{2} )}{3}

Evaluate

\frac{1}{3} e^{t} - \frac{1}{3} e^{- \frac{t}{2}} cos (\frac{3 t}{2}) + \frac{e ^{- \frac{t}{2}} sin ( \frac{3 t}{2} )}{3}

at point

0 = 0

= \frac{e ^{t} + 2 e ^{- \frac{t}{2}} cos ( \frac{3 t}{2} )}{3}

2 e^{t} + 5

\int 1 + 2 x^{2} + x^{4} d x

\frac{1}{x + 4}, 2

3 e^{\frac{1}{x}}

\int_{0}^{π} sin^{2} (8 x) d x

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