What is inverse oflaplace 5/(s+1) ?

The answer to inverse oflaplace 5/(s+1) is 5e^{-t}

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inverse oflaplace 5/(s+1)

inverse laplace

\frac{5}{s + 1}

5 e^{- t}

Solution steps

L^{- 1} {\frac{5}{s + 1}}

= L^{- 1} {5 \cdot \frac{1}{s + 1}}

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)}

= 5 L^{- 1} {\frac{1}{s + 1}}

Use Inverse Laplace Transform table:

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

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

= 5 e^{- t}

4 x^{4} - 4 x^{2}, 3 x^{2}, [0, 0]

n = 1 \sum \infty cos^{n} (1.6)

\frac{d}{d x} (e^{x} x^{2})

x^{7} arcsin (x)

\int (5 x^{2} + 7 x + 11) d x

What is inverse oflaplace 5/(s+1) ?
The answer to inverse oflaplace 5/(s+1) is 5e^{-t}