What is inverse oflaplace s/(s+5) ?

The answer to inverse oflaplace s/(s+5) is \delta(t)-5e^{-5t}

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inverse oflaplace s/(s+5)

inverse laplace

\frac{s}{s + 5}

δ (t) - 5 e^{- 5 t}

Solution steps

L^{- 1} {\frac{s}{s + 5}}

Take the partial fraction of

\frac{s}{s + 5} : 1 - \frac{5}{s + 5}

= L^{- 1} {1 - \frac{5}{s + 5}}

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{5}{s + 5}}

L^{- 1} {1} : δ (t)

L^{- 1} {\frac{5}{s + 5}} : 5 e^{- 5 t}

= δ (t) - 5 e^{- 5 t}

f (x) = 4 x^{3} - x^{2} - 7, at x = - 1

\int 8 tan (x) sec^{3} (x) d x

y = (x^{2} + 2)^{2} (x^{4} + 4)^{4}

\int \frac{1}{y ^{3} - y} d y

h (x) = \frac{7}{x}, at x = 5

What is inverse oflaplace s/(s+5) ?
The answer to inverse oflaplace s/(s+5) is \delta(t)-5e^{-5t}