What is inverse oflaplace ((s+1))/((s^2+4s+20)) ?

The answer to inverse oflaplace ((s+1))/((s^2+4s+20)) is e^{-2t}cos(4t)-1/4 e^{-2t}sin(4t)

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inverse oflaplace ((s+1))/((s^2+4s+20))

inverse laplace

\frac{( s + 1 )}{( s ^{2} + 4 s + 20 )}

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

Solution steps

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

Expand

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

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

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

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

Refine

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

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

x \to - \infty lim (\frac{4 + 5 e ^{4 x}}{5 - 6 e ^{5 x}})

\int - 2 x 1 - x^{2} d x

\frac{d}{d x} y - \frac{1}{cos ( x )} = 0

f (x) = ln (x + 1 + x^{2})

x \to 8 - lim (\frac{4}{x - 8})

What is inverse oflaplace ((s+1))/((s^2+4s+20)) ?
The answer to inverse oflaplace ((s+1))/((s^2+4s+20)) is e^{-2t}cos(4t)-1/4 e^{-2t}sin(4t)