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

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

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

inverse laplace

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

e^{- t} cos (t) + 2 e^{- t} sin (t)

Solution steps

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

Expand

\frac{s + 3}{s ^{2} + 2 s + 2} : \frac{s + 1}{( s + 1 ) ^{2} + 1} + 2 \cdot \frac{1}{( s + 1 ) ^{2} + 1}

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

5 x^{2} + 8 x - 8

\frac{d}{d u} (u)

\frac{d}{d x} ((6 x - 1) (5 x - 2)^{- 1})

\frac{d}{d x} (2 (x - 1)^{2})

y^{^{''}} + 4 y^{^{'}} + 5 y = - 10 x + 3 e^{- x}

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