What is laplacetransform (t-3)^2 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

laplacetransform (t-3)^2

laplace transform

(t - 3)^{2}

\frac{2}{s ^{3}} - \frac{6}{s ^{2}} + \frac{9}{s}

Solution steps

L {(t - 3)^{2}}

Expand

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

= L {t^{2} - 6 t + 9}

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}} - 6 L {t} + L {9}

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

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

L {9} : \frac{9}{s}

= \frac{2}{s ^{3}} - 6 \cdot \frac{1}{s ^{2}} + \frac{9}{s}

Refine

\frac{2}{s ^{3}} - 6 \frac{1}{s ^{2}} + \frac{9}{s} : \frac{2}{s ^{3}} - \frac{6}{s ^{2}} + \frac{9}{s}

= \frac{2}{s ^{3}} - \frac{6}{s ^{2}} + \frac{9}{s}

n \to \infty lim (5^{- n})

\int 3 x^{2} - 1 d x

\frac{d}{d x} (e^{x + 3})

\int_{0}^{1} \frac{- 10 y}{e ^{2 y}} d y

x \to 3 + lim (\frac{( x - 8 )}{x - 3})

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