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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse laplace

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

2 H (t) + t - 2 e^{t} + e^{t} t

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

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

\int (\frac{x}{2}) d x

\int \frac{6 x}{( x ^{2} + 7 ) ^{2}} d x

\int 6 cos (t) d t

\int \frac{y}{1 + y ^{2}} d y

\int 8 (x + 1) ln (x + 1) d x

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