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

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

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inverse oflaplace 4/(s^2+4)

inverse laplace

\frac{4}{s ^{2} + 4}

2 sin (2 t)

Solution steps

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

= L^{- 1} {2 \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)}

= 2 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)

= 2 sin (2 t)

n = 0 \sum \infty 3^{n} x^{n}

\int_{- 2}^{3} (8 x^{3} + 3 x - 1) d x

\int e^{x} - \frac{1}{2} y^{2} d x

- 2 x^{\frac{3}{2}}

\frac{d y}{d x} = \frac{1 - x - y}{x + y}

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