What is laplacetransform (t-3)^3 ?

The answer to laplacetransform (t-3)^3 is 6/(s^4)-(18}{s^3}+(27)/(s^2)-\frac{27)/s

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laplacetransform (t-3)^3

laplace transform

(t - 3)^{3}

\frac{6}{s ^{4}} - \frac{18}{s ^{3}} + \frac{27}{s ^{2}} - \frac{27}{s}

Solution steps

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

Expand

(t - 3)^{3} : t^{3} - 9 t^{2} + 27 t - 27

= L {t^{3} - 9 t^{2} + 27 t - 27}

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}} - 9 L {t^{2}} + 27 L {t} - L {27}

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

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

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

L {27} : \frac{27}{s}

= \frac{6}{s ^{4}} - 9 \cdot \frac{2}{s ^{3}} + 27 \cdot \frac{1}{s ^{2}} - \frac{27}{s}

Refine

\frac{6}{s ^{4}} - 9 \frac{2}{s ^{3}} + 27 \frac{1}{s ^{2}} - \frac{27}{s} : \frac{6}{s ^{4}} - \frac{18}{s ^{3}} + \frac{27}{s ^{2}} - \frac{27}{s}

= \frac{6}{s ^{4}} - \frac{18}{s ^{3}} + \frac{27}{s ^{2}} - \frac{27}{s}

\int_{0}^{2} \frac{x}{( 1 + 2 x ^{2} ) ^{2}} d x

\int_{- 18}^{32} \frac{1}{2 x} d x

\frac{d ^{2}}{d x ^{2}} (x^{2} + 11)

\int_{0}^{1} x^{2} e^{x^{3} + 0} d x

y^{^{''}} + y = x

What is laplacetransform (t-3)^3 ?
The answer to laplacetransform (t-3)^3 is 6/(s^4)-(18}{s^3}+(27)/(s^2)-\frac{27)/s