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

The answer to inverse oflaplace s/(s^2-2s+4) is e^tcos(sqrt(3)t)+1/(sqrt(3))e^tsin(sqrt(3)t)

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

inverse laplace

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

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

Solution steps

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

Expand

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

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

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

L^{- 1} {\frac{s - 1}{( s - 1 ) ^{2} + 3}} : e^{t} cos (3 t)

L^{- 1} {\frac{1}{( s - 1 ) ^{2} + 3}} : e^{t} \frac{1}{3} sin (3 t)

= e^{t} cos (3 t) + 1 \cdot e^{t} \frac{1}{3} sin (3 t)

Refine

e^{t} cos (3 t) + 1 e^{t} \frac{1}{3} sin (3 t) : e^{t} cos (3 t) + \frac{1}{3} e^{t} sin (3 t)

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

\frac{\partial}{\partial u} (3 sin (v) cos (u))

\frac{d}{d x} (\frac{3}{4 + x ^{2}})

\frac{( s - 2 )}{s ^{2} ( s - 1 ) ^{2}}

sin (t) cos (x) d t - sin (x) cos (t) d x = 0

y^{2} = x, x = 2 y

What is inverse oflaplace s/(s^2-2s+4) ?
The answer to inverse oflaplace s/(s^2-2s+4) is e^tcos(sqrt(3)t)+1/(sqrt(3))e^tsin(sqrt(3)t)