What is laplacetransform (t-5)^3 ?

The answer to laplacetransform (t-5)^3 is 6/(s^4)-(30}{s^3}+(75)/(s^2)-\frac{125)/s

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laplacetransform (t-5)^3

laplace transform

(t - 5)^{3}

\frac{6}{s ^{4}} - \frac{30}{s ^{3}} + \frac{75}{s ^{2}} - \frac{125}{s}

Solution steps

L {(t - 5)^{3}}

Expand

(t - 5)^{3} : t^{3} - 15 t^{2} + 75 t - 125

= L {t^{3} - 15 t^{2} + 75 t - 125}

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}} - 15 L {t^{2}} + 75 L {t} - L {125}

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

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

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

L {125} : \frac{125}{s}

= \frac{6}{s ^{4}} - 15 \cdot \frac{2}{s ^{3}} + 75 \cdot \frac{1}{s ^{2}} - \frac{125}{s}

Refine

\frac{6}{s ^{4}} - 15 \frac{2}{s ^{3}} + 75 \frac{1}{s ^{2}} - \frac{125}{s} : \frac{6}{s ^{4}} - \frac{30}{s ^{3}} + \frac{75}{s ^{2}} - \frac{125}{s}

= \frac{6}{s ^{4}} - \frac{30}{s ^{3}} + \frac{75}{s ^{2}} - \frac{125}{s}

\frac{d}{d x} (\frac{5 x ^{2}}{2})

(y) d x - 4 (x + y^{6}) d y = 0

f (x) = 3 - 4 x^{2}, (4, - 61)

5 cos (t)

\frac{d}{d x} (3 e^{x} sin (x))

What is laplacetransform (t-5)^3 ?
The answer to laplacetransform (t-5)^3 is 6/(s^4)-(30}{s^3}+(75)/(s^2)-\frac{125)/s