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

The answer to inverse oflaplace (5s)/(s^2+4s+5) is 5e^{-2t}cos(t)-10e^{-2t}sin(t)

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

inverse laplace

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

5 e^{- 2 t} cos (t) - 10 e^{- 2 t} sin (t)

Solution steps

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

Expand

\frac{5 s}{s ^{2} + 4 s + 5} : 5 \cdot \frac{s + 2}{( s + 2 ) ^{2} + 1} - 10 \cdot \frac{1}{( s + 2 ) ^{2} + 1}

= L^{- 1} {5 \cdot \frac{s + 2}{( s + 2 ) ^{2} + 1} - 10 \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)}

= 5 L^{- 1} {\frac{s + 2}{( s + 2 ) ^{2} + 1}} - 10 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)

= 5 e^{- 2 t} cos (t) - 10 e^{- 2 t} sin (t)

(2 y + 2 t - 9) d t + (4 y + 2 t - 1) d y = 0

\int (3 - x)^{2} d x

f^{'} (x) = f (x) (1 - f (x))

\frac{d}{d t} (2 t^{3})

\frac{d ^{3}}{d x ^{3}} (4 x + 8)

What is inverse oflaplace (5s)/(s^2+4s+5) ?
The answer to inverse oflaplace (5s)/(s^2+4s+5) is 5e^{-2t}cos(t)-10e^{-2t}sin(t)