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

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

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

inverse laplace

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

e^{t} \cdot 3 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

= e^{t} \cdot 3 t

\int \frac{48 x ^{2}}{( x - 15 ) ( x + 5 ) ^{2}} d x

x \to - 3 lim (4 x^{2} + 3 x + 2)

(1 - x^{2}) \frac{d y}{d x} = x y + y

y = \frac{2 x + 1}{x}

\frac{d}{d t} (- 6 t)

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