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)

x \to a lim (\frac{ln ( x ) - ln ( a )}{x - a})

\int (\frac{sec ( x )}{tan ( x )})^{4} d x

\frac{d y}{d x} = e^{4 x - 3 y}, y (0) = 2

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

y = x

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)