What is inverse oflaplace (2s+9)/(s^2+4s+40) ?

The answer to inverse oflaplace (2s+9)/(s^2+4s+40) is 2e^{-2t}cos(6t)+5/6 e^{-2t}sin(6t)

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inverse oflaplace (2s+9)/(s^2+4s+40)

inverse laplace

\frac{2 s + 9}{s ^{2} + 4 s + 40}

2 e^{- 2 t} cos (6 t) + \frac{5}{6} e^{- 2 t} sin (6 t)

Solution steps

L^{- 1} {\frac{2 s + 9}{s ^{2} + 4 s + 40}}

Expand

\frac{2 s + 9}{s ^{2} + 4 s + 40} : 2 \cdot \frac{s + 2}{( s + 2 ) ^{2} + 36} + 5 \cdot \frac{1}{( s + 2 ) ^{2} + 36}

= L^{- 1} {2 \cdot \frac{s + 2}{( s + 2 ) ^{2} + 36} + 5 \cdot \frac{1}{( s + 2 ) ^{2} + 36}}

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} + 36}} + 5 L^{- 1} {\frac{1}{( s + 2 ) ^{2} + 36}}

L^{- 1} {\frac{s + 2}{( s + 2 ) ^{2} + 36}} : e^{- 2 t} cos (6 t)

L^{- 1} {\frac{1}{( s + 2 ) ^{2} + 36}} : e^{- 2 t} \frac{1}{6} sin (6 t)

= 2 e^{- 2 t} cos (6 t) + 5 e^{- 2 t} \frac{1}{6} sin (6 t)

Refine

2 e^{- 2 t} cos (6 t) + 5 e^{- 2 t} \frac{1}{6} sin (6 t) : 2 e^{- 2 t} cos (6 t) + \frac{5}{6} e^{- 2 t} sin (6 t)

= 2 e^{- 2 t} cos (6 t) + \frac{5}{6} e^{- 2 t} sin (6 t)

\frac{\partial}{\partial t} (\frac{t}{c ^{2} t ^{2} - x ^{2}})

y^{^{''}} + 12 y^{^{'}} + 36 = 0

\frac{\partial}{\partial x} (y e^{x} + x ln (z))

t y^{^{'}} = 4 t^{2} - 2 y

y^{^{''}} x = y^{^{'}}

What is inverse oflaplace (2s+9)/(s^2+4s+40) ?
The answer to inverse oflaplace (2s+9)/(s^2+4s+40) is 2e^{-2t}cos(6t)+5/6 e^{-2t}sin(6t)