What is laplacetransform (1+e^t)^2 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

laplacetransform (1+e^t)^2

laplace transform

(1 + e^{t})^{2}

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

Solution steps

L {(1 + e^{t})^{2}}

Expand

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

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

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 {1} + 2 L {e^{t}} + L {e^{2 t}}

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

L {e^{t}} : \frac{1}{s - 1}

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

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

Refine

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

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

\frac{\partial}{\partial x} (8 + x ln (x y - 5))

\int \frac{x ^{2}}{u} d u

f (x) = x^{2} arctan (4 x)

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

h (x) = \frac{7}{x ^{2}}

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