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

The answer to inverse oflaplace 4/(s^2+2s+4) is 4/(sqrt(3))e^{-t}sin(sqrt(3)t)

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

inverse laplace

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

\frac{4}{3} e^{- t} sin (3 t)

Solution steps

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

Expand

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

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

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

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

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

= 4 e^{- t} \frac{1}{3} sin (3 t)

Refine

4 e^{- t} \frac{1}{3} sin (3 t) : \frac{4}{3} e^{- t} sin (3 t)

= \frac{4}{3} e^{- t} sin (3 t)

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

\frac{\partial}{\partial s} (- (s^{2} + t^{2}))

\int_{1}^{e} ln (13 x) d x

\int e^{1 - 8 t} d t

\frac{\partial}{\partial u} (\frac{1}{4} (u - v))

What is inverse oflaplace 4/(s^2+2s+4) ?
The answer to inverse oflaplace 4/(s^2+2s+4) is 4/(sqrt(3))e^{-t}sin(sqrt(3)t)