What is laplacetransform (t-8)^3 ?

The answer to laplacetransform (t-8)^3 is 6/(s^4)-(48}{s^3}+(192)/(s^2)-\frac{512)/s

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laplacetransform (t-8)^3

laplace transform

(t - 8)^{3}

\frac{6}{s ^{4}} - \frac{48}{s ^{3}} + \frac{192}{s ^{2}} - \frac{512}{s}

Solution steps

L {(t - 8)^{3}}

Expand

(t - 8)^{3} : t^{3} - 24 t^{2} + 192 t - 512

= L {t^{3} - 24 t^{2} + 192 t - 512}

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)}

= L {t^{3}} - 24 L {t^{2}} + 192 L {t} - L {512}

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

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

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

L {512} : \frac{512}{s}

= \frac{6}{s ^{4}} - 24 \cdot \frac{2}{s ^{3}} + 192 \cdot \frac{1}{s ^{2}} - \frac{512}{s}

Refine

\frac{6}{s ^{4}} - 24 \frac{2}{s ^{3}} + 192 \frac{1}{s ^{2}} - \frac{512}{s} : \frac{6}{s ^{4}} - \frac{48}{s ^{3}} + \frac{192}{s ^{2}} - \frac{512}{s}

= \frac{6}{s ^{4}} - \frac{48}{s ^{3}} + \frac{192}{s ^{2}} - \frac{512}{s}

\frac{d y}{d x} = \frac{3 x ^{2} + 4 x - 4}{2 y - 4}, y (1) = 3

n = 3 \sum \infty \frac{1}{2} \cdot (\frac{5}{7})^{n}

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

\int x^{0.5} d x

\int \frac{1}{x ^{3} + 4 x} d x

What is laplacetransform (t-8)^3 ?
The answer to laplacetransform (t-8)^3 is 6/(s^4)-(48}{s^3}+(192)/(s^2)-\frac{512)/s