What is inverse oflaplace 2/(s+3) ?

The answer to inverse oflaplace 2/(s+3) is 2e^{-3t}

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inverse oflaplace 2/(s+3)

inverse laplace

\frac{2}{s + 3}

2 e^{- 3 t}

Solution steps

L^{- 1} {\frac{2}{s + 3}}

= L^{- 1} {2 \cdot \frac{1}{s + 3}}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= 2 L^{- 1} {\frac{1}{s + 3}}

Use Inverse Laplace Transform table:

L^{- 1} {\frac{1}{s - a}} = e^{a t}

L^{- 1} {\frac{1}{s + 3}} = e^{- 3 t}

= 2 e^{- 3 t}

\frac{s + 1}{( s ^{2} + 9 )}

f (x) = 2 x^{2} - 3 x - 2, at x = 6

\int_{- \infty}^{- 1} e^{- 4 t} d t

f (x) = (x^{2} - 4) (3 x + 1)

f (x) = \frac{( e ^{x} )}{7 x ^{2} + 1}

What is inverse oflaplace 2/(s+3) ?
The answer to inverse oflaplace 2/(s+3) is 2e^{-3t}