What is inverse oflaplace 1/(2s^3+6s^2+4s) ?

The answer to inverse oflaplace 1/(2s^3+6s^2+4s) is 1/4 \H(t)-1/2 e^{-t}+1/4 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/(2s^3+6s^2+4s)

inverse laplace

\frac{1}{2 s ^{3} + 6 s ^{2} + 4 s}

\frac{1}{4} H (t) - \frac{1}{2} e^{- t} + \frac{1}{4} e^{- 2 t}

Solution steps

L^{- 1} {\frac{1}{2 s ^{3} + 6 s ^{2} + 4 s}}

Take the partial fraction of

\frac{1}{2 s ^{3} + 6 s ^{2} + 4 s} : \frac{1}{4 s} - \frac{1}{2 ( s + 1 )} + \frac{1}{4 ( s + 2 )}

= L^{- 1} {\frac{1}{4 s} - \frac{1}{2 ( s + 1 )} + \frac{1}{4 ( 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}{4 s}} - L^{- 1} {\frac{1}{2 ( s + 1 )}} + L^{- 1} {\frac{1}{4 ( s + 2 )}}

L^{- 1} {\frac{1}{4 s}} : \frac{1}{4} H (t)

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

L^{- 1} {\frac{1}{4 ( s + 2 )}} : \frac{1}{4} e^{- 2 t}

= \frac{1}{4} H (t) - \frac{1}{2} e^{- t} + \frac{1}{4} e^{- 2 t}

\int x (x) d x

y = (3 x + 4)^{5} + 15 x (3 x + 4)^{4}

\int - e^{4 x} d x

n = 1 \sum \infty \frac{n}{csc ( \frac{1}{n} )}

\int \frac{( arctan ( x ) ) ^{4}}{1 + x ^{2}} d x

What is inverse oflaplace 1/(2s^3+6s^2+4s) ?
The answer to inverse oflaplace 1/(2s^3+6s^2+4s) is 1/4 \H(t)-1/2 e^{-t}+1/4 e^{-2t}