What is inverse oflaplace 2/(s+8) ?

The answer to inverse oflaplace 2/(s+8) is 2e^{-8t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 2/(s+8)

inverse laplace

\frac{2}{s + 8}

2 e^{- 8 t}

Solution steps

L^{- 1} {\frac{2}{s + 8}}

= L^{- 1} {2 \cdot \frac{1}{s + 8}}

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 + 8}}

Use Inverse Laplace Transform table:

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

L^{- 1} {\frac{1}{s + 8}} = e^{- 8 t}

= 2 e^{- 8 t}

(- 4, 2), (- 1, - 5)

x \to 3 - lim (\frac{∣ x ∣}{x})

\int 5 x - 4 d x

\int x ln (x + 4) d x

\int ln (y) + 2 x y^{3} d x

What is inverse oflaplace 2/(s+8) ?
The answer to inverse oflaplace 2/(s+8) is 2e^{-8t}