What is laplacetransform 2(t-2)^2+7(t-2)+9 ?

The answer to laplacetransform 2(t-2)^2+7(t-2)+9 is 4/(s^3)-1/(s^2)+3/s

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

laplacetransform 2(t-2)^2+7(t-2)+9

laplace transform

2 (t - 2)^{2} + 7 (t - 2) + 9

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

Solution steps

L {2 (t - 2)^{2} + 7 (t - 2) + 9}

Expand

2 (t - 2)^{2} + 7 (t - 2) + 9 : 2 t^{2} - t + 3

= L {2 t^{2} - t + 3}

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)}

= 2 L {t^{2}} - L {t} + L {3}

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

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

L {3} : \frac{3}{s}

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

Refine

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

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

\frac{d}{d x} (sin (2 x^{3} + x^{2} + 1))

x \to 0 lim (\frac{a x - a ln ( x + 1 )}{x ln ( x + 1 )})

\int_{2}^{10} \frac{1}{x} d x

\frac{d}{d x} (x e^{(2 x)})

\frac{\partial}{\partial x} (9 x^{2} + y^{2})

What is laplacetransform 2(t-2)^2+7(t-2)+9 ?
The answer to laplacetransform 2(t-2)^2+7(t-2)+9 is 4/(s^3)-1/(s^2)+3/s