What is laplacetransform f(t)=(1+e^{2t})^2 ?

The answer to laplacetransform f(t)=(1+e^{2t})^2 is 1/s+2/(s-2)+1/(s-4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

laplacetransform f(t)=(1+e^{2t})^2

laplace transform

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

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

Solution steps

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

Expand

(1 + e^{2 t})^{2} : 1 + 2 e^{2 t} + e^{4 t}

= L {1 + 2 e^{2 t} + e^{4 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^{2 t}} + L {e^{4 t}}

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

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

L {e^{4 t}} : \frac{1}{s - 4}

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

Refine

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

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

\frac{d}{d x} (4 - 6 x)

\frac{d}{d x} (x - 9)

\int 14 arctan (x) d x

\frac{d x}{d t} + t x^{3} + \frac{x}{t} = 0

What is laplacetransform f(t)=(1+e^{2t})^2 ?
The answer to laplacetransform f(t)=(1+e^{2t})^2 is 1/s+2/(s-2)+1/(s-4)