What is inverse oflaplace (20)/(s(s^2+2s+5)) ?

The answer to inverse oflaplace (20)/(s(s^2+2s+5)) is 4\H(t)-4e^{-t}cos(2t)-2e^{-t}sin(2t)

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inverse oflaplace (20)/(s(s^2+2s+5))

inverse laplace

\frac{20}{s ( s ^{2} + 2 s + 5 )}

4 H (t) - 4 e^{- t} cos (2 t) - 2 e^{- t} sin (2 t)

Solution steps

L^{- 1} {\frac{20}{s ( s ^{2} + 2 s + 5 )}}

Take the partial fraction of

\frac{20}{s ( s ^{2} + 2 s + 5 )} : \frac{4}{s} + \frac{- 4 s - 8}{s ^{2} + 2 s + 5}

= L^{- 1} {\frac{4}{s} + \frac{- 4 s - 8}{s ^{2} + 2 s + 5}}

Expand

\frac{- 4 s - 8}{s ^{2} + 2 s + 5} : - 4 \cdot \frac{s + 1}{( s + 1 ) ^{2} + 4} - 4 \cdot \frac{1}{( s + 1 ) ^{2} + 4}

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

L^{- 1} {\frac{4}{s}} : 4 H (t)

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)

= 4 H (t) - 4 e^{- t} cos (2 t) - 4 e^{- t} \frac{1}{2} sin (2 t)

Refine

4 H (t) - 4 e^{- t} cos (2 t) - 4 e^{- t} \frac{1}{2} sin (2 t) : 4 H (t) - 4 e^{- t} cos (2 t) - 2 e^{- t} sin (2 t)

= 4 H (t) - 4 e^{- t} cos (2 t) - 2 e^{- t} sin (2 t)

x \to 8 - lim (\frac{x + 7}{x + 8})

\int \frac{1}{4 - ( x - 1 ) ^{2}} d x

\frac{d y}{d x} y = (5 - x)^{633}

\int_{0}^{2} t e^{- t} d t

\frac{x}{x + 11}

What is inverse oflaplace (20)/(s(s^2+2s+5)) ?
The answer to inverse oflaplace (20)/(s(s^2+2s+5)) is 4\H(t)-4e^{-t}cos(2t)-2e^{-t}sin(2t)