What is inverse oflaplace 1/(x^2+0.0184) ?

The answer to inverse oflaplace 1/(x^2+0.0184) is 7.37209…sin(0.13564…t)

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inverse oflaplace 1/(x^2+0.0184)

inverse laplace

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

7.37209 \dots sin (0.13564 \dots t)

Solution steps

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

= L^{- 1} {\frac{1}{0.13564 \dots} \cdot \frac{0.13564 \dots}{x ^{2} + 0.13564 \dots ^{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}{0.13564 \dots} L^{- 1} {\frac{0.13564 \dots}{x ^{2} + 0.13564 \dots ^{2}}}

Use Inverse Laplace Transform table:

L^{- 1} {\frac{a}{s ^{2} + a ^{2}}} = sin (a t)

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

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

Simplify

\frac{1}{0.13564 \dots} sin (0.13564 \dots t) : 7.37209 \dots sin (0.13564 \dots t)

= 7.37209 \dots sin (0.13564 \dots t)

\int \frac{10}{( x - 1 ) ( x ^{2} - 9 )} d x

x \to 0 - lim (8 - x^{2})

f (x) = 1 + cos (x), at x = \frac{π}{2}

\frac{d x}{d t} + t^{5} x^{7} + \frac{x}{t} = 0

f (x) = x cos (x)

What is inverse oflaplace 1/(x^2+0.0184) ?
The answer to inverse oflaplace 1/(x^2+0.0184) is 7.37209…sin(0.13564…t)