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

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

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

inverse laplace

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

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

Solution steps

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

Expand

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

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

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

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

\int_{t}^{2} \frac{1}{x - 1} d x

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

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

cos (sin (x))

\frac{\partial}{\partial x} (e^{\frac{- 2 y}{x}})

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