What is inverse oflaplace s/(s^2+4s+13) ?

The answer to inverse oflaplace s/(s^2+4s+13) is e^{-2t}cos(3t)-2/3 e^{-2t}sin(3t)

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inverse oflaplace s/(s^2+4s+13)

inverse laplace

\frac{s}{s ^{2} + 4 s + 13}

e^{- 2 t} cos (3 t) - \frac{2}{3} e^{- 2 t} sin (3 t)

Solution steps

L^{- 1} {\frac{s}{s ^{2} + 4 s + 13}}

Expand

\frac{s}{s ^{2} + 4 s + 13} : \frac{s + 2}{( s + 2 ) ^{2} + 9} - 2 \cdot \frac{1}{( s + 2 ) ^{2} + 9}

= L^{- 1} {\frac{s + 2}{( s + 2 ) ^{2} + 9} - 2 \cdot \frac{1}{( s + 2 ) ^{2} + 9}}

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 + 2}{( s + 2 ) ^{2} + 9}} - 2 L^{- 1} {\frac{1}{( s + 2 ) ^{2} + 9}}

L^{- 1} {\frac{s + 2}{( s + 2 ) ^{2} + 9}} : e^{- 2 t} cos (3 t)

L^{- 1} {\frac{1}{( s + 2 ) ^{2} + 9}} : e^{- 2 t} \frac{1}{3} sin (3 t)

= e^{- 2 t} cos (3 t) - 2 e^{- 2 t} \frac{1}{3} sin (3 t)

Refine

e^{- 2 t} cos (3 t) - 2 e^{- 2 t} \frac{1}{3} sin (3 t) : e^{- 2 t} cos (3 t) - \frac{2}{3} e^{- 2 t} sin (3 t)

= e^{- 2 t} cos (3 t) - \frac{2}{3} e^{- 2 t} sin (3 t)

x \to 4.2 lim (3)

\int e^{- s t} sin (a t) d t

\int_{- 7}^{9} (\frac{x}{2} + 9) d x

x \to 2 lim (\frac{x}{x})

\frac{d}{d x} (cos (2 x) \cdot 2)

What is inverse oflaplace s/(s^2+4s+13) ?
The answer to inverse oflaplace s/(s^2+4s+13) is e^{-2t}cos(3t)-2/3 e^{-2t}sin(3t)