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

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

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

inverse laplace

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

e^{t} + e^{t} t

Solution steps

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

Take the partial fraction of

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

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

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

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

= e^{t} + e^{t} t

\int x^{2} sin (2 x) d x

\frac{d}{d x} ((a x^{2} + b x + c) e^{\frac{- x}{6}})

x \to 7 lim (\frac{5 x}{x + 2})

t \to - 2 lim (t^{2} - 6 t - 16)

y^{^{''}} - 2 y^{^{'}} + y = e^{t} arctan (t)

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