What is inverse oflaplace (s+1)/((s^2+2)(s+1)) ?

The answer to inverse oflaplace (s+1)/((s^2+2)(s+1)) is 1/(sqrt(2))sin(sqrt(2)t)

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inverse oflaplace (s+1)/((s^2+2)(s+1))

inverse laplace

\frac{s + 1}{( s ^{2} + 2 ) ( s + 1 )}

\frac{1}{2} sin (2 t)

Solution steps

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

Expand

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

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

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

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

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

Refine

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

= \frac{1}{2} sin (2 t)

9 sin (2 x)

f (x) = \frac{x}{4 + x ^{2}}

(4^{x})^{^{'}}

n = 1 \sum \infty ln (\frac{n}{n + 1})

y^{^{''}} - 6 y^{^{'}} + 11 y = 0

What is inverse oflaplace (s+1)/((s^2+2)(s+1)) ?
The answer to inverse oflaplace (s+1)/((s^2+2)(s+1)) is 1/(sqrt(2))sin(sqrt(2)t)