What is inverse oflaplace 2/(4s^2-8s) ?

The answer to inverse oflaplace 2/(4s^2-8s) is -1/4 \H(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 2/(4s^2-8s)

inverse laplace

\frac{2}{4 s ^{2} - 8 s}

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

Solution steps

L^{- 1} {\frac{2}{4 s ^{2} - 8 s}}

Take the partial fraction of

\frac{2}{4 s ^{2} - 8 s} : - \frac{1}{4 s} + \frac{1}{4 ( s - 2 )}

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

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

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

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

\int_{2}^{4} \frac{( 2 x - 3 ) x}{6} d x

x \to - 2 lim (- x^{2} + 6 x - 1)

f (x) = 2 x + \frac{3}{x}

x \to 4 lim (\frac{5}{x + 2})

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

What is inverse oflaplace 2/(4s^2-8s) ?
The answer to inverse oflaplace 2/(4s^2-8s) is -1/4 \H(t)+1/4 e^{2t}