What is laplacetransform 2*((t-1)^2)*1((t-1)) ?

The answer to laplacetransform 2*((t-1)^2)*1((t-1)) is (12)/(s^4)-(12)/(s^3)+6/(s^2)-2/s

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laplacetransform 2((t-1)^2)1((t-1))

laplace transform

2 \cdot ((t - 1)^{2}) \cdot 1 ((t - 1))

\frac{12}{s ^{4}} - \frac{12}{s ^{3}} + \frac{6}{s ^{2}} - \frac{2}{s}

Solution steps

L {2 \cdot 1 \cdot ((t - 1)^{2}) ((t - 1))}

Expand

2 (t - 1)^{2} 1 (t - 1) : 2 t^{3} - 6 t^{2} + 6 t - 2

= L {2 t^{3} - 6 t^{2} + 6 t - 2}

Use the linearity property of Laplace Transform:

For functions

f (t), g (t)

and constants

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= 2 L {t^{3}} - 6 L {t^{2}} + 6 L {t} - L {2}

L {t^{3}} : \frac{6}{s ^{4}}

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

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

L {2} : \frac{2}{s}

= 2 \cdot \frac{6}{s ^{4}} - 6 \cdot \frac{2}{s ^{3}} + 6 \cdot \frac{1}{s ^{2}} - \frac{2}{s}

Refine

2 \frac{6}{s ^{4}} - 6 \frac{2}{s ^{3}} + 6 \frac{1}{s ^{2}} - \frac{2}{s} : \frac{12}{s ^{4}} - \frac{12}{s ^{3}} + \frac{6}{s ^{2}} - \frac{2}{s}

= \frac{12}{s ^{4}} - \frac{12}{s ^{3}} + \frac{6}{s ^{2}} - \frac{2}{s}

\frac{\partial}{\partial x} (\frac{7 x}{3 y})

f (x) = \frac{x}{2 x - 3}, (1, - 1)

\int \frac{e ^{- s t}}{s ^{2}} d t

y = \frac{100}{x ^{4}}

f (x) = 14400 + 600 x + x^{2}

What is laplacetransform 2*((t-1)^2)*1((t-1)) ?
The answer to laplacetransform 2*((t-1)^2)*1((t-1)) is (12)/(s^4)-(12)/(s^3)+6/(s^2)-2/s