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

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

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

inverse laplace

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

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

Solution steps

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

Expand

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

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

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

= 5 e^{- t} cos (2 t) - 5 e^{- t} \frac{1}{2} sin (2 t)

Refine

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

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

\int 3 cos^{2} (x) sin^{2} (x) d x

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

\int_{1}^{2} \frac{4 + u ^{2}}{u ^{3}} d u

\frac{d y}{d x} = \frac{x ^{2}}{y}

\frac{d x}{d t} = 2 (x^{2} + 1), x (\frac{π}{4}) = 1

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