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

The answer to inverse oflaplace (-3s-4)/(2(s^2+2s+2)) is -3/2 e^{-t}cos(t)-1/2 e^{-t}sin(t)

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

inverse laplace

\frac{- 3 s - 4}{2 ( s ^{2} + 2 s + 2 )}

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

Solution steps

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

Expand

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

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

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

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

\frac{e ^{- 8 s}}{s}

\frac{d}{d x} (e^{2 e^{x}})

\int ((csc (y))^{2} - sec (y) tan (y)) d y

\int_{- π}^{π} x^{2} sin (x) d x

(4 x y + 3 y^{2} - x) d x + x (x + 2 y) d y = 0

What is inverse oflaplace (-3s-4)/(2(s^2+2s+2)) ?
The answer to inverse oflaplace (-3s-4)/(2(s^2+2s+2)) is -3/2 e^{-t}cos(t)-1/2 e^{-t}sin(t)