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

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

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

inverse laplace

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

- 2 sin (2 t) + 2 sin (t)

Solution steps

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

Take the partial fraction of

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

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

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

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

= - 2 sin (2 t) + 2 sin (t)

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