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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

Refine

- \frac{1}{4} e^{- 2 t} + \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{d}{d x} (2 e^{3 x^{2} + 6 x + 7})

\int cos (3 x - 2) d x

f (x) = \frac{1}{x + 1}, at x = 1

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

\int \frac{7}{x ^{2} ( x ^{2} + 25 )} d x

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