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

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

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inverse oflaplace s/(1+s)

inverse laplace

\frac{s}{1 + s}

δ (t) - e^{- t}

Solution steps

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

Take the partial fraction of

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

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

L^{- 1} {1} : δ (t)

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

= δ (t) - e^{- t}

\int \frac{1}{484 + x ^{2}} d x

5 \frac{d y}{d x} + 45 y = 9

\frac{d}{d x} (x^{2} cos (y) y^{^{'}})

f (x) = 2 cos (3 x)

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