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

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

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laplacetransform (t-1)1(t-1)

laplace transform

(t - 1) 1 (t - 1)

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

Solution steps

L {1 \cdot (t - 1) (t - 1)}

Expand

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

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

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

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

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

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

Refine

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

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

\int tan (θ) sec^{2} (θ) d θ

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

\int ln (x^{2}) d x

\frac{x d y}{d x} + y = e^{- x}

9 y^{^{''}} + 6 y^{^{'}} + 6 y = 0

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