What is inverse oflaplace 2/(s(5s+1)) ?

The answer to inverse oflaplace 2/(s(5s+1)) is 2\H(t)-2e^{-t/5}

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inverse oflaplace 2/(s(5s+1))

inverse laplace

\frac{2}{s ( 5 s + 1 )}

2 H (t) - 2 e^{- \frac{t}{5}}

Solution steps

L^{- 1} {\frac{2}{s ( 5 s + 1 )}}

Take the partial fraction of

\frac{2}{s ( 5 s + 1 )} : \frac{2}{s} - \frac{10}{5 s + 1}

= L^{- 1} {\frac{2}{s} - \frac{10}{5 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{10}{5 s + 1}}

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

L^{- 1} {\frac{10}{5 s + 1}} : 2 e^{- \frac{t}{5}}

= 2 H (t) - 2 e^{- \frac{t}{5}}

e^{- y^{2}}

\int + 3 x^{3} e^{- x^{2}} d x

\int (8) d x

\int_{0}^{π} \int 1 d x d y

x \to \infty lim (x^{- 5})

What is inverse oflaplace 2/(s(5s+1)) ?
The answer to inverse oflaplace 2/(s(5s+1)) is 2\H(t)-2e^{-t/5}