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

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

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inverse oflaplace 1/(5s+1)

inverse laplace

\frac{1}{5 s + 1}

\frac{1}{5} e^{- \frac{t}{5}}

Solution steps

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

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

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)}

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

Use Inverse Laplace Transform table:

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

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

= \frac{1}{5} e^{- \frac{t}{5}}

\int \cdot 1 0^{- 3} + \cdot 1 0^{- 5} t + \cdot 1 0^{- 8} t^{2} + \cdot 1 0^{- 12} t^{3} d t

\int (9 x^{2} + 25) (\frac{7 x - 4}{x ^{2}}) d x

\frac{d}{d x} (e^{5 x} + 5 e^{5 x} x)

\frac{d}{d x} (x^{2} + 4 x y + y^{6} - 8)

\int_{3}^{4} 9 x d x

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