What is inverse oflaplace (3-2s)/(s^2+4s+5) ?

The answer to inverse oflaplace (3-2s)/(s^2+4s+5) is -2e^{-2t}cos(t)+7e^{-2t}sin(t)

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inverse oflaplace (3-2s)/(s^2+4s+5)

inverse laplace

\frac{3 - 2 s}{s ^{2} + 4 s + 5}

- 2 e^{- 2 t} cos (t) + 7 e^{- 2 t} sin (t)

Solution steps

L^{- 1} {\frac{3 - 2 s}{s ^{2} + 4 s + 5}}

Expand

\frac{3 - 2 s}{s ^{2} + 4 s + 5} : - 2 \cdot \frac{s + 2}{( s + 2 ) ^{2} + 1} + 7 \cdot \frac{1}{( s + 2 ) ^{2} + 1}

= L^{- 1} {- 2 \cdot \frac{s + 2}{( s + 2 ) ^{2} + 1} + 7 \cdot \frac{1}{( s + 2 ) ^{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)}

= - 2 L^{- 1} {\frac{s + 2}{( s + 2 ) ^{2} + 1}} + 7 L^{- 1} {\frac{1}{( s + 2 ) ^{2} + 1}}

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

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

= - 2 e^{- 2 t} cos (t) + 7 e^{- 2 t} sin (t)

\frac{\partial}{\partial x} (ln (2 x + 2))

\frac{d}{d x} (- \frac{3}{2 ( 1 + x ) ^{\frac{3}{2}}})

\frac{d}{d x} (1 ln (x) - \frac{x ^{2}}{8})

g (u) = 6 u + 5 u

\int \frac{4}{5} - \frac{1}{3 5 x} d x

What is inverse oflaplace (3-2s)/(s^2+4s+5) ?
The answer to inverse oflaplace (3-2s)/(s^2+4s+5) is -2e^{-2t}cos(t)+7e^{-2t}sin(t)