What is inverse oflaplace 1/(s+1)*1/s ?

The answer to inverse oflaplace 1/(s+1)*1/s is \H(t)-e^{-t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse oflaplace 1/(s+1)*1/s

inverse laplace

\frac{1}{s + 1} \cdot \frac{1}{s}

H (t) - e^{- t}

Solution steps

L^{- 1} {\frac{1}{s + 1} \cdot \frac{1}{s}}

Expand

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

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

Take the partial fraction of

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

= L^{- 1} {\frac{1}{s} - \frac{1}{s + 1}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{1}{s}} - L^{- 1} {\frac{1}{s + 1}}

L^{- 1} {\frac{1}{s}} : H (t)

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

= H (t) - e^{- t}

n = 0 \sum \infty n (\frac{1}{4})^{n}

x \to 0 lim (- 6 + 6 x^{3})

y = 3^{2} + 3

\int e^{2 a x} d x

\int_{- 5}^{4} t^{2} d t

What is inverse oflaplace 1/(s+1)*1/s ?
The answer to inverse oflaplace 1/(s+1)*1/s is \H(t)-e^{-t}