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}

y (1 + x^{2}) y^{^{'}} - x (6 + y^{2}) = 0

y = 5 x^{4} e^{x} + e^{x} x^{5}

x \to 1 lim (- 6 + 2 x^{2})

f (x) = \frac{x + 2}{x ^{2}}

36 y^{^{''}} - y = 0, y (- 6) = 1, y^{^{'}} (- 6) = - 1

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