What is inverse oflaplace 1/(s^2+8)(-40/33) ?

The answer to inverse oflaplace 1/(s^2+8)(-40/33) is -(10sqrt(2)sin(2sqrt(2)t))/(33)

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inverse oflaplace 1/(s^2+8)(-40/33)

inverse laplace

\frac{1}{s ^{2} + 8} (- \frac{40}{33})

- \frac{10 2 sin ( 2 2 t )}{33}

Solution steps

L^{- 1} {(- \frac{40}{33}) \frac{1}{s ^{2} + 8}}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= (- \frac{40}{33}) L^{- 1} {\frac{1}{s ^{2} + 8}}

L^{- 1} {\frac{1}{s ^{2} + 8}} : \frac{1}{2 2} sin (22 t)

= - \frac{40}{33} \cdot \frac{1}{2 2} sin (22 t)

Refine

- \frac{40}{33} \frac{1}{2 2} sin (22 t) : - \frac{10 2 sin ( 2 2 t )}{33}

= - \frac{10 2 sin ( 2 2 t )}{33}

What is inverse oflaplace 1/(s^2+8)(-40/33) ?
The answer to inverse oflaplace 1/(s^2+8)(-40/33) is -(10sqrt(2)sin(2sqrt(2)t))/(33)