What is laplacetransform (t-1)^4 ?

The answer to laplacetransform (t-1)^4 is (24)/(s^5)-(24)/(s^4)+(12)/(s^3)-4/(s^2)+1/s

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laplacetransform (t-1)^4

laplace transform

(t - 1)^{4}

\frac{24}{s ^{5}} - \frac{24}{s ^{4}} + \frac{12}{s ^{3}} - \frac{4}{s ^{2}} + \frac{1}{s}

Solution steps

L {(t - 1)^{4}}

Expand

(t - 1)^{4} : t^{4} - 4 t^{3} + 6 t^{2} - 4 t + 1

= L {t^{4} - 4 t^{3} + 6 t^{2} - 4 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)}

= L {t^{4}} - 4 L {t^{3}} + 6 L {t^{2}} - 4 L {t} + L {1}

L {t^{4}} : \frac{24}{s ^{5}}

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}

= \frac{24}{s ^{5}} - 4 \cdot \frac{6}{s ^{4}} + 6 \cdot \frac{2}{s ^{3}} - 4 \cdot \frac{1}{s ^{2}} + \frac{1}{s}

Refine

\frac{24}{s ^{5}} - 4 \frac{6}{s ^{4}} + 6 \frac{2}{s ^{3}} - 4 \frac{1}{s ^{2}} + \frac{1}{s} : \frac{24}{s ^{5}} - \frac{24}{s ^{4}} + \frac{12}{s ^{3}} - \frac{4}{s ^{2}} + \frac{1}{s}

= \frac{24}{s ^{5}} - \frac{24}{s ^{4}} + \frac{12}{s ^{3}} - \frac{4}{s ^{2}} + \frac{1}{s}

\frac{d}{d x} (\frac{2 x ^{4} + 5}{4 - 4 x})

\frac{\partial}{\partial x} (25 - x^{2})

\frac{d y}{d x} + \frac{1}{x + 8} y = x^{- 2}

\frac{x ^{2} - 2 x}{( x - 1 ) ^{2}}

f (x) = x^{4} + 9 e^{x}, (0, 9)

What is laplacetransform (t-1)^4 ?
The answer to laplacetransform (t-1)^4 is (24)/(s^5)-(24)/(s^4)+(12)/(s^3)-4/(s^2)+1/s