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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace (-1)/(s+1)

inverse laplace

\frac{- 1}{s + 1}

- e^{- t}

Solution steps

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

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

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 + 1}}

Use Inverse Laplace Transform table:

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

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

= - 1 \cdot e^{- t}

= - e^{- t}

\int \frac{x - 3}{x ^{3} - 2 x ^{2} + 3 x} d x

x \to 3 lim (\frac{1}{2 ^{x} - 8})

x \to 0 lim (\frac{x}{tan ( 5 x )})

\int \frac{sin ( 2 x )}{cos ^{2} ( 2 x )} d x

x \to 1 lim (x + 2)

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