What is inverse oflaplace (0.07s)/(7s^2+19980) ?

The answer to inverse oflaplace (0.07s)/(7s^2+19980) is 0.01cos(53.42551…t)

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inverse oflaplace (0.07s)/(7s^2+19980)

inverse laplace

\frac{0.07 s}{7 s ^{2} + 19980}

0.01 cos (53.42551 \dots t)

Solution steps

L^{- 1} {\frac{0.07 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{0.07 s}{7 s ^{2} + 19980} : F (s) = \frac{0.07}{7 s ^{2} + 19980}

= \frac{d}{d t} (L^{- 1} {\frac{0.07}{7 s ^{2} + 19980}}) + L^{- 1} {\frac{0.07}{7 s ^{2} + 19980}} (0)

L^{- 1} {\frac{0.07}{7 s ^{2} + 19980}} : \frac{0.07}{6 3885} sin (\frac{6 555 t}{7})

\frac{d}{d t} (\frac{0.07}{6 3885} sin (\frac{6 555 t}{7})) = 0.01 cos (53.42551 \dots t)

Evaluate

\frac{0.07}{6 3885} sin (\frac{6 555 t}{7})

at point

0 = 0

= 0.01 cos (53.42551 \dots t)

e^{- 7 x}, x \geq 0

\frac{d}{d x} (\frac{e ^{x}}{2 x ^{2} + 1})

\frac{d y}{d x} + x^{2} y = 10 x^{2}, y (1) = 5

\int_{- \infty}^{0} \frac{1}{5 - 2 x} d x

\frac{3 x}{7 x ^{2} + 1}

What is inverse oflaplace (0.07s)/(7s^2+19980) ?
The answer to inverse oflaplace (0.07s)/(7s^2+19980) is 0.01cos(53.42551…t)