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

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

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inverse oflaplace 6/(s^2+4)

inverse laplace

\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_{2}^{4} (2 - x^{2}) - (10 - 6 x) d x

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{2 x}

\frac{\partial}{\partial h} (\frac{A}{h + 5 h})

x \to 0 lim (\frac{sec ( 2 x ) - 1}{7 x})

\int x 3 1 - x d x

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