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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 1/(s(s+2))

inverse laplace

\frac{1}{s ( s + 2 )}

\frac{1}{2} H (t) - \frac{1}{2} e^{- 2 t}

Solution steps

L^{- 1} {\frac{1}{s ( s + 2 )}}

Take the partial fraction of

\frac{1}{s ( s + 2 )} : \frac{1}{2 s} - \frac{1}{2 ( s + 2 )}

= L^{- 1} {\frac{1}{2 s} - \frac{1}{2 ( s + 2 )}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{1}{2 s}} - L^{- 1} {\frac{1}{2 ( s + 2 )}}

L^{- 1} {\frac{1}{2 s}} : \frac{1}{2} H (t)

L^{- 1} {\frac{1}{2 ( s + 2 )}} : \frac{1}{2} e^{- 2 t}

= \frac{1}{2} H (t) - \frac{1}{2} e^{- 2 t}

\int 13 (tan (x)) (ln (cos (x))) d x

x \to + \frac{π}{2} lim (cot (x))

f (x) = 5 arcsin (x^{3})

\frac{e ^{x}}{x}

\int_{1}^{2} x ln (x) d x

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