What is inverse oflaplace 8/(3s^2+12) ?

The answer to inverse oflaplace 8/(3s^2+12) is 4/3 sin(2t)

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inverse oflaplace 8/(3s^2+12)

inverse laplace

\frac{8}{3 s ^{2} + 12}

\frac{4}{3} sin (2 t)

Solution steps

L^{- 1} {\frac{8}{3 s ^{2} + 12}}

= L^{- 1} {\frac{4}{3} \cdot \frac{2}{s ^{2} + 2 ^{2}}}

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{4}{3} L^{- 1} {\frac{2}{s ^{2} + 2 ^{2}}}

Use Inverse Laplace Transform table:

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

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

= \frac{4}{3} sin (2 t)

\int tan^{3} (6 x) sec^{10} (6 x) d x

\frac{d ^{2}}{d x ^{2}} (x (2 x + 3)^{5})

f (x) = \frac{3 x ^{2} + 1}{2 x}

h \to 0 lim (3 - 4 h)

\frac{d}{d x} (- 600 x^{- 2})

What is inverse oflaplace 8/(3s^2+12) ?
The answer to inverse oflaplace 8/(3s^2+12) is 4/3 sin(2t)