What is inverse oflaplace (2s+12)/((s^2+2s+5)) ?

The answer to inverse oflaplace (2s+12)/((s^2+2s+5)) is 2e^{-t}cos(2t)+5e^{-t}sin(2t)

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inverse oflaplace (2s+12)/((s^2+2s+5))

inverse laplace

\frac{2 s + 12}{( s ^{2} + 2 s + 5 )}

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

Solution steps

L^{- 1} {\frac{2 s + 12}{( s ^{2} + 2 s + 5 )}}

Expand

\frac{2 s + 12}{s ^{2} + 2 s + 5} : 2 \cdot \frac{s + 1}{( s + 1 ) ^{2} + 4} + 10 \cdot \frac{1}{( s + 1 ) ^{2} + 4}

= L^{- 1} {2 \cdot \frac{s + 1}{( s + 1 ) ^{2} + 4} + 10 \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)}

= 2 L^{- 1} {\frac{s + 1}{( s + 1 ) ^{2} + 4}} + 10 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)

= 2 e^{- t} cos (2 t) + 10 e^{- t} \frac{1}{2} sin (2 t)

Refine

2 e^{- t} cos (2 t) + 10 e^{- t} \frac{1}{2} sin (2 t) : 2 e^{- t} cos (2 t) + 5 e^{- t} sin (2 t)

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

\frac{d x}{d t} = sin (t)

x \to \infty lim (5 - x)

\frac{d}{d x} (sin^{2} (π x))

\int x^{- 8} d x

y = \frac{x}{1 + x ^{2}}, at x = 0

What is inverse oflaplace (2s+12)/((s^2+2s+5)) ?
The answer to inverse oflaplace (2s+12)/((s^2+2s+5)) is 2e^{-t}cos(2t)+5e^{-t}sin(2t)