What is inverse oflaplace (10)/(5s+1) ?

The answer to inverse oflaplace (10)/(5s+1) is 2e^{-t/5}

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inverse oflaplace (10)/(5s+1)

inverse laplace

\frac{10}{5 s + 1}

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

Solution steps

L^{- 1} {\frac{10}{5 s + 1}}

= L^{- 1} {2 \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)}

= 2 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}}

= 2 e^{- \frac{t}{5}}

\int 4 sin (x) + \frac{2 x ^{5} - x}{x} d x

f (x) = - 3 x^{4} + 7 x^{2} - 4, at x = 1

\frac{\partial}{\partial x} (2 x - 5 y)

\frac{1}{s ^{2} + 0.1 s + 1}

\int (n x)^{\frac{( n - 1 )}{n}} d x

What is inverse oflaplace (10)/(5s+1) ?
The answer to inverse oflaplace (10)/(5s+1) is 2e^{-t/5}