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}

\int (1 - 8 x) (ln (x) + 1) d x

6 x^{2} + x y + 6 y^{2} = 13, (1, 1)

y^{^{''}} + 9 y = 5 csc^{2} (3 t)

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