What is laplacetransform 4e^{-t}+2 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

laplacetransform 4e^{-t}+2

laplace transform

4 e^{- t} + 2

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

Solution steps

L {4 e^{- t} + 2}

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)}

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

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

L {2} : \frac{2}{s}

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

Refine

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

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

\int \frac{1}{1 + v ^{2}} d v

\int \frac{( ln ( x ) ) ^{5}}{x} d x

\frac{d}{d x} (5 sec (5 x) tan (5 x))

\int_{0}^{0.5} 8 y^{2} d y

\frac{5}{t ^{2}}

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