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)

x \to \infty lim (csc (x))

3 \frac{d y}{d x} = x^{2}

\int \frac{x + 8}{3 x ^{2} - 12} d x

\int (\frac{1}{x ( ln ( x ) - 1 )}) d x

\int \frac{lo g _{3} ( x )}{3 x} d x

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