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

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

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

inverse laplace

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

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

Solution steps

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

Expand

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

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

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

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

Refine

- \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}{2} sin (2 t)

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

y = - x^{2} + 2 x + 2, y = x - 4

\int \frac{d ^{2} y}{1} d x^{2}

\int (sin^{2} (x) + cos^{2} (x) - 1) d x

\int - \frac{6 x}{5} d x

y = (x^{3} - 25 x)^{14}

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