What is inverse oflaplace F(s)= 6/(s^2+4) ?

The answer to inverse oflaplace F(s)= 6/(s^2+4) is 3sin(2t)

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inverse oflaplace F(s)= 6/(s^2+4)

inverse laplace

F (s) = \frac{6}{s ^{2} + 4}

3 sin (2 t)

Solution steps

L^{- 1} {\frac{6}{s ^{2} + 4}}

= L^{- 1} {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)}

= 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)

= 3 sin (2 t)

\int - 6 sin (2 x) + 3 cos (3 x) d x

\frac{d y}{d x} = \frac{15 x}{y}

\frac{d}{d y} (e^{x e^{y}})

x, x - 2, y = 0

What is inverse oflaplace F(s)= 6/(s^2+4) ?
The answer to inverse oflaplace F(s)= 6/(s^2+4) is 3sin(2t)