What is laplacetransform f(t)=(2t-1)^3 ?

The answer to laplacetransform f(t)=(2t-1)^3 is (48)/(s^4)-(24)/(s^3)+6/(s^2)-1/s

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laplacetransform f(t)=(2t-1)^3

laplace transform

f (t) = (2 t - 1)^{3}

\frac{48}{s ^{4}} - \frac{24}{s ^{3}} + \frac{6}{s ^{2}} - \frac{1}{s}

Solution steps

L {(2 t - 1)^{3}}

Expand

(2 t - 1)^{3} : 8 t^{3} - 12 t^{2} + 6 t - 1

= L {8 t^{3} - 12 t^{2} + 6 t - 1}

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)}

= 8 L {t^{3}} - 12 L {t^{2}} + 6 L {t} - L {1}

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

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

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

L {1} : \frac{1}{s}

= 8 \cdot \frac{6}{s ^{4}} - 12 \cdot \frac{2}{s ^{3}} + 6 \cdot \frac{1}{s ^{2}} - \frac{1}{s}

Refine

8 \frac{6}{s ^{4}} - 12 \frac{2}{s ^{3}} + 6 \frac{1}{s ^{2}} - \frac{1}{s} : \frac{48}{s ^{4}} - \frac{24}{s ^{3}} + \frac{6}{s ^{2}} - \frac{1}{s}

= \frac{48}{s ^{4}} - \frac{24}{s ^{3}} + \frac{6}{s ^{2}} - \frac{1}{s}

2 y^{^{'}} = y^{2}

\frac{\partial}{\partial x} (y (- sin (x ln (y))) ln (y))

\frac{d}{d t} (- \frac{1}{t ^{2}})

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

\frac{d}{d x} ((A x^{2} + B x + C) e^{x})

What is laplacetransform f(t)=(2t-1)^3 ?
The answer to laplacetransform f(t)=(2t-1)^3 is (48)/(s^4)-(24)/(s^3)+6/(s^2)-1/s