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

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

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inverse oflaplace 5/(5s+1)

inverse laplace

\frac{5}{5 s + 1}

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

Solution steps

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

= L^{- 1} {1 \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)}

= 1 \cdot 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}}

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

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

x \to - 3 lim (\frac{x ^{2} + 1}{2 x})

f (x) = 4 x^{- \frac{1}{2}}

\frac{d}{d h} (1.18 h)

\frac{\partial}{\partial x} (\frac{( 7 x - 8 y )}{1 x + 2 y})

\frac{d}{d x} (\frac{cos ( x ^{2} )}{x ^{2} + 1})

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