What is inverse oflaplace (-3)/((s-1)^2) ?

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

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inverse oflaplace (-3)/((s-1)^2)

inverse laplace

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

- 3 e^{t} t

Solution steps

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{- 3}{( s - 1 ) ^{2}} : a = 1, F (s) = \frac{- 3}{s ^{2}}

= e^{t} L^{- 1} {\frac{- 3}{s ^{2}}}

L^{- 1} {\frac{- 3}{s ^{2}}} : - 3 t

= - 3 e^{t} t

f (x) = \frac{x ^{3}}{3} + 4 x^{2} + 12 x + 19

\int - x cos (x) d x

\frac{\partial}{\partial x} (A e^{- 3 x})

y^{^{''}} - 4 y^{^{'}} + 13 y = 0, y (0) = - 1, y^{^{'}} (0) = - 5

f (x) = arccos (e^{2 x})

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