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

Popular >

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}

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

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

\int \frac{e ^{x}}{3 + 5 e ^{2 x}} d x

\frac{d}{d t} (cos (π t))

\frac{\partial}{\partial x} (x^{3} - 3 x y^{2} + y^{3})

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