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

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

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

inverse laplace

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

δ (t) - \frac{1}{2} e^{- \frac{t}{2}}

Solution steps

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

Take the partial fraction of

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

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

L^{- 1} {1} : δ (t)

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

= δ (t) - \frac{1}{2} e^{- \frac{t}{2}}

\frac{d}{d x} (\frac{4}{x ^{3}} + 3 x)

6 cos (6 x)

\int_{- 1}^{0} x^{2} e^{x^{3} + 1} d x

y^{5} \cdot (y^{^{'}} - y) = e^{6 x}

\frac{d y}{d x} = - \frac{2 y ^{2}}{x ^{2}}

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