What is inverse oflaplace (1/500)/(140+8s+s^2) ?

The answer to inverse oflaplace (1/500)/(140+8s+s^2) is 1/(1000sqrt(31))e^{-4t}sin(2sqrt(31)t)

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inverse oflaplace (1/500)/(140+8s+s^2)

inverse laplace

\frac{\frac{1}{500}}{140 + 8 s + s ^{2}}

\frac{1}{1000 31} e^{- 4 t} sin (231 t)

Solution steps

L^{- 1} {\frac{\frac{1}{500}}{140 + 8 s + s ^{2}}}

Expand

\frac{\frac{1}{500}}{140 + 8 s + s ^{2}} : \frac{1}{500} \cdot \frac{1}{( s + 4 ) ^{2} + 124}

= L^{- 1} {\frac{1}{500} \cdot \frac{1}{( s + 4 ) ^{2} + 124}}

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}{500} L^{- 1} {\frac{1}{( s + 4 ) ^{2} + 124}}

L^{- 1} {\frac{1}{( s + 4 ) ^{2} + 124}} : e^{- 4 t} \frac{1}{2 31} sin (231 t)

= \frac{1}{500} e^{- 4 t} \frac{1}{2 31} sin (231 t)

Refine

\frac{1}{500} e^{- 4 t} \frac{1}{2 31} sin (231 t) : \frac{1}{1000 31} e^{- 4 t} sin (231 t)

= \frac{1}{1000 31} e^{- 4 t} sin (231 t)

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

y^{^{''}} + 6.2832 y = 10

y = x^{2} (x + 3)

x \to \infty lim (1.9)

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

What is inverse oflaplace (1/500)/(140+8s+s^2) ?
The answer to inverse oflaplace (1/500)/(140+8s+s^2) is 1/(1000sqrt(31))e^{-4t}sin(2sqrt(31)t)