What is laplacetransform (t-4)^2 ?

The answer to laplacetransform (t-4)^2 is 2/(s^3)-8/(s^2)+(16)/s

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laplacetransform (t-4)^2

laplace transform

(t - 4)^{2}

\frac{2}{s ^{3}} - \frac{8}{s ^{2}} + \frac{16}{s}

Solution steps

L {(t - 4)^{2}}

Expand

(t - 4)^{2} : t^{2} - 8 t + 16

= L {t^{2} - 8 t + 16}

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^{2}} - 8 L {t} + L {16}

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

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

L {16} : \frac{16}{s}

= \frac{2}{s ^{3}} - 8 \cdot \frac{1}{s ^{2}} + \frac{16}{s}

Refine

\frac{2}{s ^{3}} - 8 \frac{1}{s ^{2}} + \frac{16}{s} : \frac{2}{s ^{3}} - \frac{8}{s ^{2}} + \frac{16}{s}

= \frac{2}{s ^{3}} - \frac{8}{s ^{2}} + \frac{16}{s}

\frac{\partial}{\partial y} (x^{3})

(x^{4} + y^{4}) d x - x y^{3} d y = 0

\frac{d y}{d x} - tan (x) - \frac{1}{x} = 0

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

\int_{0}^{1} \frac{x ^{3} - 5 x - 9}{x ^{2} - x - 6} d x

What is laplacetransform (t-4)^2 ?
The answer to laplacetransform (t-4)^2 is 2/(s^3)-8/(s^2)+(16)/s