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)

\int x^{2} - 7 x + 5 d x

\int_{\frac{9}{8}}^{4} \frac{1}{1 - x ^{2}} d x

\int - \frac{x}{1 + x ^{2}} d x

\int \frac{1}{3 x 9 x ^{2} - 12} d x

e^{- 1}

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)