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

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

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

inverse laplace

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

δ (t) - \frac{1}{2} e^{- \frac{t}{2}}

Solution steps

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

Take the partial fraction of

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

\int \frac{x ^{4} + 2}{x ^{2}} d x

f (x) = x - 3, (7, 2)

\frac{d}{d x} (- \frac{6 x}{y})

\frac{9}{x}

x \to - 1 lim (\frac{1}{( x + 1 ) ^{2}})

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