What is inverse oflaplace (0.001s)/(s^2+4) ?

The answer to inverse oflaplace (0.001s)/(s^2+4) is 0.001cos(2t)

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inverse oflaplace (0.001s)/(s^2+4)

inverse laplace

\frac{0.001 s}{s ^{2} + 4}

0.001 cos (2 t)

Solution steps

L^{- 1} {\frac{0.001 s}{s ^{2} + 4}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

For

\frac{0.001 s}{s ^{2} + 4} : F (s) = \frac{0.001}{s ^{2} + 4}

= \frac{d}{d t} (L^{- 1} {\frac{0.001}{s ^{2} + 4}}) + L^{- 1} {\frac{0.001}{s ^{2} + 4}} (0)

L^{- 1} {\frac{0.001}{s ^{2} + 4}} : \frac{0.001}{2} sin (2 t)

\frac{d}{d t} (\frac{0.001}{2} sin (2 t)) = 0.001 cos (2 t)

Evaluate

\frac{0.001}{2} sin (2 t)

at point

0 = 0

= 0.001 cos (2 t)

\int \frac{13 x - 6}{x ( 3 x - 2 )} d x

n = 1 \sum \infty \frac{( 1 ^{n} )}{n}

\frac{d}{d x} ((2 x) d x)

\int \frac{ln ( y )}{y} d y

\frac{d}{d x} (x^{2} e^{2})

What is inverse oflaplace (0.001s)/(s^2+4) ?
The answer to inverse oflaplace (0.001s)/(s^2+4) is 0.001cos(2t)