What is inverse oflaplace (0.01s)/(7s^2+19980) ?

The answer to inverse oflaplace (0.01s)/(7s^2+19980) is 0.00142…cos(53.42551…t)

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inverse oflaplace (0.01s)/(7s^2+19980)

inverse laplace

\frac{0.01 s}{7 s ^{2} + 19980}

0.00142 \dots cos (53.42551 \dots t)

Solution steps

L^{- 1} {\frac{0.01 s}{7 s ^{2} + 19980}}

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{0.01 s}{7 s ^{2} + 19980} : F (s) = \frac{0.01}{7 s ^{2} + 19980}

= \frac{d}{d t} (L^{- 1} {\frac{0.01}{7 s ^{2} + 19980}}) + L^{- 1} {\frac{0.01}{7 s ^{2} + 19980}} (0)

L^{- 1} {\frac{0.01}{7 s ^{2} + 19980}} : \frac{0.01}{6 3885} sin (\frac{6 555 t}{7})

\frac{d}{d t} (\frac{0.01}{6 3885} sin (\frac{6 555 t}{7})) = 0.00142 \dots cos (53.42551 \dots t)

Evaluate

\frac{0.01}{6 3885} sin (\frac{6 555 t}{7})

at point

0 = 0

= 0.00142 \dots cos (53.42551 \dots t)

\frac{d}{d x} (\frac{- ln ( 1 - 6 x )}{x})

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

\frac{d}{d x} (7 x y)

\frac{d}{d x} (arcsec (2 x + 1))

\int x 1 + 3 x d x

What is inverse oflaplace (0.01s)/(7s^2+19980) ?
The answer to inverse oflaplace (0.01s)/(7s^2+19980) is 0.00142…cos(53.42551…t)