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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

\int \frac{cos ( x )}{1 + ( sin ( x ) ) ^{2}} d x

\int \frac{1}{x 25 x ^{2} + 49} d x

\frac{1}{2 s ^{2} + 6 s + 4}

\frac{d}{d x} (\frac{2 ( x ^{2} - 9 )}{x ^{2} - 4})

n = 0 \sum \infty \frac{n !}{8 ^{n}}

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