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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

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

f (x) = \frac{1}{1 - x}, at x = 2

\frac{d y}{d t} = y - 4 t + (y - 4 t)^{2} + 4

\int x (2 x^{2} + 1)^{5} d x

(- 3, 5), (2, - 1)

x \to - 2 - lim (\frac{3}{x ^{2} - 4})

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