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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse laplace

\frac{1}{( 5 \cdot 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}}

\frac{d}{d x} ((x + 1)^{4})

\frac{d}{d x} (\frac{a x}{x ^{2} + a ^{2}})

\int (x^{4} + 2 x - 16) d x

\frac{d}{d x} (e^{x^{3}})

\frac{d}{d x} (a (1 - cos (x)))

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