What is laplacetransform (t-1)(3(t-1)^2+9(t-1)+3) ?

The answer to laplacetransform (t-1)(3(t-1)^2+9(t-1)+3) is (18)/(s^4)-6/(s^2)+3/s

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laplacetransform (t-1)(3(t-1)^2+9(t-1)+3)

laplace transform

(t - 1) (3 (t - 1)^{2} + 9 (t - 1) + 3)

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

Solution steps

L {(t - 1) (3 (t - 1)^{2} + 9 (t - 1) + 3)}

Expand

(t - 1) (3 (t - 1)^{2} + 9 (t - 1) + 3) : 3 t^{3} - 6 t + 3

= L {3 t^{3} - 6 t + 3}

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)}

= 3 L {t^{3}} - 6 L {t} + L {3}

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

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

L {3} : \frac{3}{s}

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

Refine

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

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

\int \frac{1}{sec ( y )} d y

x \to 0 lim (x (sin (\frac{1}{x})))

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

\frac{\partial}{\partial x} (ln (y^{2} + x^{2}))

n = 0 \sum \infty \frac{2 n}{2 ^{n}}

What is laplacetransform (t-1)(3(t-1)^2+9(t-1)+3) ?
The answer to laplacetransform (t-1)(3(t-1)^2+9(t-1)+3) is (18)/(s^4)-6/(s^2)+3/s