What is laplacetransform (t-2)^2 ?

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

laplacetransform (t-2)^2

laplace transform

(t - 2)^{2}

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

Solution steps

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

Expand

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

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

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}} - 4 L {t} + L {4}

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

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

L {4} : \frac{4}{s}

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

Refine

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

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

\int_{1}^{4} \frac{x ^{2} + 8}{5 x - x ^{2}} d x

\int \frac{3. 2 ^{x} - 2. 3 ^{x}}{2 ^{x}} d x

\int \frac{42 x}{3 x ^{2} + e} d x

y^{^{'}} + y + 2 sin (x) = 0

d er i v a t i v e e^{2 x} + 2 x

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