What is laplacetransform (2t-1)^2 ?

The answer to laplacetransform (2t-1)^2 is 8/(s^3)-4/(s^2)+1/s

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laplacetransform (2t-1)^2

laplace transform

(2 t - 1)^{2}

\frac{8}{s ^{3}} - \frac{4}{s ^{2}} + \frac{1}{s}

Solution steps

L {(2 t - 1)^{2}}

Expand

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

= L {4 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)}

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

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

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

L {1} : \frac{1}{s}

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

Refine

4 \frac{2}{s ^{3}} - 4 \frac{1}{s ^{2}} + \frac{1}{s} : \frac{8}{s ^{3}} - \frac{4}{s ^{2}} + \frac{1}{s}

= \frac{8}{s ^{3}} - \frac{4}{s ^{2}} + \frac{1}{s}

3 δ (t) + 2 + 4 e^{- 3 t + 2}

\frac{\partial}{\partial x} (\frac{y}{1 + x ^{2}})

(60 + 2 t)^{\frac{2}{3}}

f (x) = 3^{4 x + 5}

y = - 1 - 8 x^{2}, (- 2, - 33)

What is laplacetransform (2t-1)^2 ?
The answer to laplacetransform (2t-1)^2 is 8/(s^3)-4/(s^2)+1/s