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

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

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

inverse laplace

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

δ (t) - e^{- 2 t}

Solution steps

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

Take the partial fraction of

\frac{s + 1}{s + 2} : 1 - \frac{1}{s + 2}

= L^{- 1} {1 - \frac{1}{s + 2}}

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 + 2}}

L^{- 1} {1} : δ (t)

L^{- 1} {\frac{1}{s + 2}} : e^{- 2 t}

= δ (t) - e^{- 2 t}

y = 3 sec (x)

\frac{d}{d x} (uv)

\int \frac{x ^{2}}{1 - x ^{3}} d x

y = 8 - x^{2}, y = x^{2}

(x^{2} + 5)^{3}

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