What is inverse oflaplace ((s+2))/(s+1) ?

The answer to inverse oflaplace ((s+2))/(s+1) is \delta(t)+e^{-t}

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inverse oflaplace ((s+2))/(s+1)

inverse laplace

\frac{( s + 2 )}{s + 1}

δ (t) + e^{- t}

Solution steps

L^{- 1} {\frac{( s + 2 )}{s + 1}}

Take the partial fraction of

\frac{s + 2}{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}

(2 y + 2 t - 3) d t + (4 y + 2 t - 1) d y = 0

\frac{\partial}{\partial x} (7 e^{x^{10} y})

x \to 5 lim (x^{2} + 2 x - 3)

f (x) = ln (4 x + 7)

\frac{d}{d t} (sin (3 t))

What is inverse oflaplace ((s+2))/(s+1) ?
The answer to inverse oflaplace ((s+2))/(s+1) is \delta(t)+e^{-t}