What is inverse oflaplace 1/(5s+2) ?

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

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inverse oflaplace 1/(5s+2)

inverse laplace

\frac{1}{5 s + 2}

\frac{1}{5} e^{- \frac{2 t}{5}}

Solution steps

L^{- 1} {\frac{1}{5 s + 2}}

= L^{- 1} {\frac{1}{5} \cdot \frac{1}{s + \frac{2}{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{2}{5}}}

Use Inverse Laplace Transform table:

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

L^{- 1} {\frac{1}{s + \frac{2}{5}}} = e^{- \frac{2 t}{5}}

= \frac{1}{5} e^{- \frac{2 t}{5}}

\frac{\partial}{\partial x} (sin^{3} (x - 5 y))

\frac{d}{d x} (30 x^{4} - 60 x^{3} + 60 x - 30)

\int (4 x^{2} + 1)^{4} + (4 x^{2} + 1)^{3} d x

\int sin (x) sin (y) d y

ln (x), x - 2, 1, 2

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