What is inverse oflaplace s/(s^2+s) ?

The answer to inverse oflaplace s/(s^2+s) is Undefined

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace s/(s^2+s)

inverse laplace

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

Undefined

Solution steps

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

For

\frac{s}{s ^{2} + s} : F (s) = \frac{1}{s ^{2} + s}

= \frac{d}{d t} (L^{- 1} {\frac{1}{s ^{2} + s}}) + L^{- 1} {\frac{1}{s ^{2} + s}} (0)

L^{- 1} {\frac{1}{s ^{2} + s}} : H (t) - e^{- t}

Evaluate

H (t) - e^{- t}

at point

0 =

Undefined

= Undefined

\frac{\partial}{\partial x} (4 x y e^{x y^{2}})

\int \frac{x ^{2}}{2} \cdot \frac{1}{x} d x

2 e^{- 5 t} sin (2 t)

y^{^{'}} = 3 3 y^{2}

f (x) = x ln (x^{2})

What is inverse oflaplace s/(s^2+s) ?
The answer to inverse oflaplace s/(s^2+s) is Undefined