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

The answer to inverse oflaplace s/(s^2+2s+4) is e^{-t}cos(sqrt(3)t)-1/(sqrt(3))e^{-t}sin(sqrt(3)t)

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

inverse laplace

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

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

Solution steps

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

Expand

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

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

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)}

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

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

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

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

Refine

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

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

\int + ln (x) d x

\frac{d y}{d x} = \frac{1}{x} (y^{2} + 0.6 y), y (1) = 0.9

x \to 4 lim (∣ x ∣ - 3^{x})

\int_{0}^{2 π} x^{2} sin (n x) d x

\frac{7}{x - 2}

What is inverse oflaplace s/(s^2+2s+4) ?
The answer to inverse oflaplace s/(s^2+2s+4) is e^{-t}cos(sqrt(3)t)-1/(sqrt(3))e^{-t}sin(sqrt(3)t)