What is inverse oflaplace s/((s+1)(s^2+4)) ?

The answer to inverse oflaplace s/((s+1)(s^2+4)) is -1/5 e^{-t}+1/5 cos(2t)+(2sin(2t))/5

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

inverse laplace

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

- \frac{1}{5} e^{- t} + \frac{1}{5} cos (2 t) + \frac{2 sin ( 2 t )}{5}

Solution steps

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

Take the partial fraction of

\frac{s}{( s + 1 ) ( s ^{2} + 4 )} : - \frac{1}{5 ( s + 1 )} + \frac{s + 4}{5 ( s ^{2} + 4 )}

= L^{- 1} {- \frac{1}{5 ( s + 1 )} + \frac{s + 4}{5 ( s ^{2} + 4 )}}

Expand

= L^{- 1} {\frac{s}{5 ( s ^{2} + 4 )} + \frac{4}{5 ( s ^{2} + 4 )} - \frac{1}{5 ( s + 1 )}}

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}{5} L^{- 1} {\frac{1}{s + 1}} + \frac{1}{5} L^{- 1} {\frac{s}{s ^{2} + 4}} + \frac{4}{5} L^{- 1} {\frac{1}{s ^{2} + 4}}

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

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

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

= - \frac{1}{5} e^{- t} + \frac{1}{5} cos (2 t) + \frac{4}{5} \cdot \frac{1}{2} sin (2 t)

Refine

- \frac{1}{5} e^{- t} + \frac{1}{5} cos (2 t) + \frac{4}{5} \frac{1}{2} sin (2 t) : - \frac{1}{5} e^{- t} + \frac{1}{5} cos (2 t) + \frac{2 sin ( 2 t )}{5}

= - \frac{1}{5} e^{- t} + \frac{1}{5} cos (2 t) + \frac{2 sin ( 2 t )}{5}

\frac{d ^{2}}{d x ^{2}} (\frac{1}{x ^{10}})

\int - tan (t) d t

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

\frac{\partial}{\partial x} (\frac{y}{x} - 1)

f (x) = cos (\frac{x ^{2} - 25}{x - 5})

What is inverse oflaplace s/((s+1)(s^2+4)) ?
The answer to inverse oflaplace s/((s+1)(s^2+4)) is -1/5 e^{-t}+1/5 cos(2t)+(2sin(2t))/5