What is inverse oflaplace 2/(s(0.12s+1)) ?

The answer to inverse oflaplace 2/(s(0.12s+1)) is 2\H(t)-2e^{-8.33333…t}

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inverse oflaplace 2/(s(0.12s+1))

inverse laplace

\frac{2}{s ( 0.12 s + 1 )}

2 H (t) - 2 e^{- 8.33333 \dots t}

Solution steps

L^{- 1} {\frac{2}{s ( 0.12 s + 1 )}}

Take the partial fraction of

\frac{2}{s ( 0.12 s + 1 )} : \frac{2}{s} - \frac{0.24}{0.12 s + 1}

= L^{- 1} {\frac{2}{s} - \frac{0.24}{0.12 s + 1}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{2}{s}} - L^{- 1} {\frac{0.24}{0.12 s + 1}}

L^{- 1} {\frac{2}{s}} : 2 H (t)

L^{- 1} {\frac{0.24}{0.12 s + 1}} : 2 e^{- 8.33333 \dots t}

= 2 H (t) - 2 e^{- 8.33333 \dots t}

x y^{^{'}} = y + 7 x^{2} sin (x)

\frac{\partial}{\partial x} (x y^{\frac{1}{2}})

R (t) = 50 t^{2} e^{- t} + 8

\frac{d}{d x} (4 x^{3} - 2 x^{2})

y^{^{''}} - 8 y^{^{'}} + 16 y = t^{- 3} e^{4 t}

What is inverse oflaplace 2/(s(0.12s+1)) ?
The answer to inverse oflaplace 2/(s(0.12s+1)) is 2\H(t)-2e^{-8.33333…t}