What is inverse oflaplace (S+1/2)/(S^2-2S+5) ?

The answer to inverse oflaplace (S+1/2)/(S^2-2S+5) is e^tcos(2t)+3/4 e^tsin(2t)

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inverse oflaplace (S+1/2)/(S^2-2S+5)

inverse laplace

\frac{S + \frac{1}{2}}{S ^{2} - 2 S + 5}

e^{t} cos (2 t) + \frac{3}{4} e^{t} sin (2 t)

Solution steps

L^{- 1} {\frac{S + \frac{1}{2}}{S ^{2} - 2 S + 5}}

Expand

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

= L^{- 1} {\frac{S - 1}{( S - 1 ) ^{2} + 4} + \frac{3}{2} \cdot \frac{1}{( S - 1 ) ^{2} + 4}}

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} + 4}} + \frac{3}{2} L^{- 1} {\frac{1}{( S - 1 ) ^{2} + 4}}

L^{- 1} {\frac{S - 1}{( S - 1 ) ^{2} + 4}} : e^{t} cos (2 t)

L^{- 1} {\frac{1}{( S - 1 ) ^{2} + 4}} : e^{t} \frac{1}{2} sin (2 t)

= e^{t} cos (2 t) + \frac{3}{2} e^{t} \frac{1}{2} sin (2 t)

Refine

e^{t} cos (2 t) + \frac{3}{2} e^{t} \frac{1}{2} sin (2 t) : e^{t} cos (2 t) + \frac{3}{4} e^{t} sin (2 t)

= e^{t} cos (2 t) + \frac{3}{4} e^{t} sin (2 t)

\frac{\partial}{\partial x} (e^{x} sin (10 y + 5 z))

\frac{d}{d x} (2 arcsin (x^{2}))

y^{^{'}} = - x (y + 1)

x \to 0 lim (∣ x + 4 ∣)

\int (3 x + 4) d x

What is inverse oflaplace (S+1/2)/(S^2-2S+5) ?
The answer to inverse oflaplace (S+1/2)/(S^2-2S+5) is e^tcos(2t)+3/4 e^tsin(2t)