What is inverse oflaplace 1/((s^2+4)) ?

The answer to inverse oflaplace 1/((s^2+4)) is 1/2 sin(2t)

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inverse oflaplace 1/((s^2+4))

inverse laplace

\frac{1}{( s ^{2} + 4 )}

\frac{1}{2} sin (2 t)

Solution steps

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

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

= \frac{1}{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)

= \frac{1}{2} sin (2 t)

\int_{- 2}^{2} (- x^{4}) - (x^{4} - 32) d x

\frac{1}{1 - x ^{2}}

f (x) = \frac{12}{x}

(4 + x) y^{^{'}} = 5 y

What is inverse oflaplace 1/((s^2+4)) ?
The answer to inverse oflaplace 1/((s^2+4)) is 1/2 sin(2t)