What is inverse oflaplace 3/(2s+1)*1/s ?

The answer to inverse oflaplace 3/(2s+1)*1/s is 3\H(t)-3e^{-t/2}

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inverse oflaplace 3/(2s+1)*1/s

inverse laplace

\frac{3}{2 s + 1} \cdot \frac{1}{s}

3 H (t) - 3 e^{- \frac{t}{2}}

Solution steps

L^{- 1} {\frac{3}{2 s + 1} \cdot \frac{1}{s}}

Expand

\frac{3}{2 s + 1} \frac{1}{s} : \frac{3}{2 s ^{2} + s}

= L^{- 1} {\frac{3}{2 s ^{2} + s}}

Take the partial fraction of

\frac{3}{2 s ^{2} + s} : \frac{3}{s} - \frac{6}{2 s + 1}

= L^{- 1} {\frac{3}{s} - \frac{6}{2 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{3}{s}} - L^{- 1} {\frac{6}{2 s + 1}}

L^{- 1} {\frac{3}{s}} : 3 H (t)

L^{- 1} {\frac{6}{2 s + 1}} : 3 e^{- \frac{t}{2}}

= 3 H (t) - 3 e^{- \frac{t}{2}}

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

t \to \frac{x}{2} lim (\frac{sin ( t )}{t})

(- 3.5) (4.3)

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

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

What is inverse oflaplace 3/(2s+1)*1/s ?
The answer to inverse oflaplace 3/(2s+1)*1/s is 3\H(t)-3e^{-t/2}