What is inverse oflaplace (3s+2)/(s^2+4s+20) ?

The answer to inverse oflaplace (3s+2)/(s^2+4s+20) is 3e^{-2t}cos(4t)-e^{-2t}sin(4t)

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inverse oflaplace (3s+2)/(s^2+4s+20)

inverse laplace

\frac{3 s + 2}{s ^{2} + 4 s + 20}

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

Solution steps

L^{- 1} {\frac{3 s + 2}{s ^{2} + 4 s + 20}}

Expand

\frac{3 s + 2}{s ^{2} + 4 s + 20} : 3 \cdot \frac{s + 2}{( s + 2 ) ^{2} + 16} - 4 \cdot \frac{1}{( s + 2 ) ^{2} + 16}

= L^{- 1} {3 \cdot \frac{s + 2}{( s + 2 ) ^{2} + 16} - 4 \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)}

= 3 L^{- 1} {\frac{s + 2}{( s + 2 ) ^{2} + 16}} - 4 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)

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

Refine

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

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

\int sin (6 x) ln (sin (6 x)) d x

\int \frac{x ^{2}}{( 36 - x ^{2} ) ^{\frac{3}{2}}} d x

\int_{0}^{π} x cos (x^{2}) d x

t^{3} y^{^{'}} + 4 t^{2} y = e^{- t}

n = 1 \sum \infty \frac{1}{( n + 1 ) !}

What is inverse oflaplace (3s+2)/(s^2+4s+20) ?
The answer to inverse oflaplace (3s+2)/(s^2+4s+20) is 3e^{-2t}cos(4t)-e^{-2t}sin(4t)