What is inverse oflaplace (s-2)/(s^2-s-6) ?

The answer to inverse oflaplace (s-2)/(s^2-s-6) is 4/5 e^{-2t}+1/5 e^{3t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace (s-2)/(s^2-s-6)

inverse laplace

\frac{s - 2}{s ^{2} - s - 6}

\frac{4}{5} e^{- 2 t} + \frac{1}{5} e^{3 t}

Solution steps

L^{- 1} {\frac{s - 2}{s ^{2} - s - 6}}

Take the partial fraction of

\frac{s - 2}{s ^{2} - s - 6} : \frac{4}{5 ( s + 2 )} + \frac{1}{5 ( s - 3 )}

= L^{- 1} {\frac{4}{5 ( s + 2 )} + \frac{1}{5 ( s - 3 )}}

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{4}{5 ( s + 2 )}} + L^{- 1} {\frac{1}{5 ( s - 3 )}}

L^{- 1} {\frac{4}{5 ( s + 2 )}} : \frac{4}{5} e^{- 2 t}

L^{- 1} {\frac{1}{5 ( s - 3 )}} : \frac{1}{5} e^{3 t}

= \frac{4}{5} e^{- 2 t} + \frac{1}{5} e^{3 t}

\frac{d}{d x} (\frac{arcsin ( 3 x )}{x})

f (x) = 3 - 4 x^{2}, at x = - 2

y = 3 x, (3, 3)

\int e^{6 - 5 t} d t

\int x sin (3 t - 3 x) d x

What is inverse oflaplace (s-2)/(s^2-s-6) ?
The answer to inverse oflaplace (s-2)/(s^2-s-6) is 4/5 e^{-2t}+1/5 e^{3t}