What is laplacetransform f(t)=(t-2)^2 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

laplacetransform f(t)=(t-2)^2

laplace transform

f (t) = (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}

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

\frac{\partial}{\partial x} (2 cos (x))

\int 3 y e^{y^{3}} d y

f (x) = ln (x - 4)

\int - cos^{2} (t) d t

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