What is inverse oflaplace 2/(s+10.04) ?

The answer to inverse oflaplace 2/(s+10.04) is 2e^{-10.04t}

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inverse oflaplace 2/(s+10.04)

inverse laplace

\frac{2}{s + 10.04}

2 e^{- 10.04 t}

Solution steps

L^{- 1} {\frac{2}{s + 10.04}}

= L^{- 1} {2 \cdot \frac{1}{s + 10.04}}

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)}

= 2 L^{- 1} {\frac{1}{s + 10.04}}

Use Inverse Laplace Transform table:

L^{- 1} {\frac{1}{s - a}} = e^{a t}

L^{- 1} {\frac{1}{s + 10.04}} = e^{- 10.04 t}

= 2 e^{- 10.04 t}

x \to - \infty lim (3 - 4 x^{2})

x \to \infty lim (\frac{x ^{3}}{2 ^{x}})

\int sin (- 2 x) d x

x \to 6 lim (6)

x \to - 1 lim (\frac{3 + x}{x ^{2} - 1})

What is inverse oflaplace 2/(s+10.04) ?
The answer to inverse oflaplace 2/(s+10.04) is 2e^{-10.04t}