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

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

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

inverse laplace

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

2 e^{- 2 t} cos (3 t) - \frac{4}{3} e^{- 2 t} sin (3 t)

Solution steps

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

Expand

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

= L^{- 1} {2 \cdot \frac{s + 2}{( s + 2 ) ^{2} + 9} - 4 \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)}

= 2 L^{- 1} {\frac{s + 2}{( s + 2 ) ^{2} + 9}} - 4 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)

= 2 e^{- 2 t} cos (3 t) - 4 e^{- 2 t} \frac{1}{3} sin (3 t)

Refine

2 e^{- 2 t} cos (3 t) - 4 e^{- 2 t} \frac{1}{3} sin (3 t) : 2 e^{- 2 t} cos (3 t) - \frac{4}{3} e^{- 2 t} sin (3 t)

= 2 e^{- 2 t} cos (3 t) - \frac{4}{3} e^{- 2 t} sin (3 t)

x \to 0 lim (sin (x) \frac{1}{x})

\int_{1}^{2} \frac{15}{x ^{3} + 2 x} d x

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

\int \frac{( ln ( z ) ) ^{5}}{z} d z

\int e^{- 12 x} d x

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