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

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

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inverse oflaplace s/((s+1))

inverse laplace

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

δ (t) - e^{- t}

Solution steps

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

Take the partial fraction of

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

\frac{\partial}{\partial z} (x^{yz})

\frac{d}{d x} (11 arccot (x) + 11 arccot (\frac{1}{x}))

\frac{- 5}{( 2 x ^{3} + 7 ) ^{2}}

(2, 7), (- 5, 2)

y^{^{''}} - 9 y^{^{'}} = 0

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