What is inverse oflaplace (6s+4)/(s^2+4s+20) ?

The answer to inverse oflaplace (6s+4)/(s^2+4s+20) is 6e^{-2t}cos(4t)-2e^{-2t}sin(4t)

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inverse oflaplace (6s+4)/(s^2+4s+20)

inverse laplace

\frac{6 s + 4}{s ^{2} + 4 s + 20}

6 e^{- 2 t} cos (4 t) - 2 e^{- 2 t} sin (4 t)

Solution steps

L^{- 1} {\frac{6 s + 4}{s ^{2} + 4 s + 20}}

Expand

\frac{6 s + 4}{s ^{2} + 4 s + 20} : 6 \cdot \frac{s + 2}{( s + 2 ) ^{2} + 16} - 8 \cdot \frac{1}{( s + 2 ) ^{2} + 16}

= L^{- 1} {6 \cdot \frac{s + 2}{( s + 2 ) ^{2} + 16} - 8 \cdot \frac{1}{( s + 2 ) ^{2} + 16}}

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)}

= 6 L^{- 1} {\frac{s + 2}{( s + 2 ) ^{2} + 16}} - 8 L^{- 1} {\frac{1}{( s + 2 ) ^{2} + 16}}

L^{- 1} {\frac{s + 2}{( s + 2 ) ^{2} + 16}} : e^{- 2 t} cos (4 t)

L^{- 1} {\frac{1}{( s + 2 ) ^{2} + 16}} : e^{- 2 t} \frac{1}{4} sin (4 t)

= 6 e^{- 2 t} cos (4 t) - 8 e^{- 2 t} \frac{1}{4} sin (4 t)

Refine

6 e^{- 2 t} cos (4 t) - 8 e^{- 2 t} \frac{1}{4} sin (4 t) : 6 e^{- 2 t} cos (4 t) - 2 e^{- 2 t} sin (4 t)

= 6 e^{- 2 t} cos (4 t) - 2 e^{- 2 t} sin (4 t)

y^{^{'}} + 5 y = e^{- 3 x}

\int sin^{6} (x) cos^{5} (x) d x

\int e^{x} + x^{e} + e + x d x

f (x) = sin (4 x) - cos (3 x)

\frac{d y}{d x} = x^{3} + x^{2} - 5 x + 11

What is inverse oflaplace (6s+4)/(s^2+4s+20) ?
The answer to inverse oflaplace (6s+4)/(s^2+4s+20) is 6e^{-2t}cos(4t)-2e^{-2t}sin(4t)