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

The answer to inverse oflaplace (s+2)/(s^2+2s+5) is e^{-t}cos(2t)+1/2 e^{-t}sin(2t)

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

inverse laplace

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

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

Solution steps

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

Expand

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

= L^{- 1} {\frac{s + 1}{( s + 1 ) ^{2} + 4} + 1 \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}} + 1 \cdot 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) + 1 \cdot e^{- t} \frac{1}{2} sin (2 t)

Refine

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

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

d er i v a t i v e 4 x + \frac{144}{x} + 80

\frac{d}{d x} (lo g_{e} (x^{2} + x + 1))

\frac{d}{d x} ((- 3 x^{5} - 3)^{3})

\int x (1 - x)^{\frac{2}{3}} d x

x \to 0 + lim (2 x^{3 x})

What is inverse oflaplace (s+2)/(s^2+2s+5) ?
The answer to inverse oflaplace (s+2)/(s^2+2s+5) is e^{-t}cos(2t)+1/2 e^{-t}sin(2t)