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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

\frac{d}{d x} (\frac{x ^{2}}{8})

y^{^{''}} + 8 y^{^{'}} + 12 y = 0

\frac{0.5}{( s ^{2} + s + 1 )} \cdot \frac{1}{s}

f (x) = π^{2}

\frac{d}{d x} ((3 x^{2} + 5)^{4})

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