What is inverse oflaplace x/(x^2+0.0184) ?

The answer to inverse oflaplace x/(x^2+0.0184) is cos(0.13564…t)

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inverse oflaplace x/(x^2+0.0184)

inverse laplace

\frac{x}{x ^{2} + 0.0184}

cos (0.13564 \dots t)

Solution steps

L^{- 1} {\frac{x}{x ^{2} + 0.0184}}

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{x}{x ^{2} + 0.0184} : F (x) = \frac{1}{x ^{2} + 0.0184}

= \frac{d}{d t} (L^{- 1} {\frac{1}{x ^{2} + 0.0184}}) + L^{- 1} {\frac{1}{x ^{2} + 0.0184}} (0)

L^{- 1} {\frac{1}{x ^{2} + 0.0184}} : \frac{1}{0.13564 \dots} sin (0.13564 \dots t)

\frac{d}{d t} (\frac{1}{0.13564 \dots} sin (0.13564 \dots t)) = cos (0.13564 \dots t)

Evaluate

\frac{1}{0.13564 \dots} sin (0.13564 \dots t)

at point

0 = 0

= cos (0.13564 \dots t)

What is inverse oflaplace x/(x^2+0.0184) ?
The answer to inverse oflaplace x/(x^2+0.0184) is cos(0.13564…t)