What is inverse oflaplace s/((s^2+4)(s+2)) ?

The answer to inverse oflaplace s/((s^2+4)(s+2)) is 1/4 cos(2t)+(sin(2t))/4-1/4 e^{-2t}

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inverse oflaplace s/((s^2+4)(s+2))

inverse laplace

\frac{s}{( s ^{2} + 4 ) ( s + 2 )}

\frac{1}{4} cos (2 t) + \frac{sin ( 2 t )}{4} - \frac{1}{4} e^{- 2 t}

Solution steps

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

Take the partial fraction of

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

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

Expand

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

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)}

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

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

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

L^{- 1} {\frac{1}{s + 2}} : e^{- 2 t}

= \frac{1}{4} cos (2 t) + \frac{2}{4} \cdot \frac{1}{2} sin (2 t) - \frac{1}{4} e^{- 2 t}

Refine

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

= \frac{1}{4} cos (2 t) + \frac{sin ( 2 t )}{4} - \frac{1}{4} e^{- 2 t}

\int - 4 x - 1 d x

y = sin (x^{5})

\int \frac{x ^{2} + 8 x - 7}{x - 3} d x

\int cos^{4} (x) - sin^{4} (x) d x

\int 4 x e^{7 x} d x

What is inverse oflaplace s/((s^2+4)(s+2)) ?
The answer to inverse oflaplace s/((s^2+4)(s+2)) is 1/4 cos(2t)+(sin(2t))/4-1/4 e^{-2t}