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

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

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

inverse laplace

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

- 3 H (t) - 2 t + 3 e^{t} - e^{t} t

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

= - 3 H (t) - 2 t + 3 e^{t} - e^{t} t

sin (t) cos (x) d t - sin (x) cos (t) d x = 0

y^{2} = x, x = 2 y

\frac{d}{d x} (vx)

\int_{0}^{π} t e^{ik t} d t

n = 1 \sum \infty \frac{n}{n + n ^{2}}

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