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

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

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

inverse laplace

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

e^{- t} cos (t) + e^{- t} sin (t)

Solution steps

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

Expand

\frac{s + 2}{s ^{2} + 2 s + 2} : \frac{s + 1}{( s + 1 ) ^{2} + 1} + 1 \cdot \frac{1}{( s + 1 ) ^{2} + 1}

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

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

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

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

= e^{- t} cos (t) + 1 \cdot e^{- t} sin (t)

Refine

e^{- t} cos (t) + 1 e^{- t} sin (t) : e^{- t} cos (t) + e^{- t} sin (t)

= e^{- t} cos (t) + e^{- t} sin (t)

\int_{0}^{1} x \frac{2 ( x + 2 )}{5} d x

f (x) = x (x - 4)^{3}

\int e^{- s t} e^{- 3 t} d t

y = x, y = 14 x, x = 25

\frac{d y}{d x} = \frac{1}{x} (y^{2} + y)

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