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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

\int x^{2} ln (9 x) d x

\frac{d}{d x} (\frac{tan ^{2} ( x )}{tan ( x ^{2} )})

x \frac{d y}{d x} = e^{7 y + x}

n = 1 \sum \infty 5 (- \frac{1}{6})^{5 n}

f (x) = \frac{2 x ^{2} - x - 1}{2 x + 2}

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