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

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

inverse laplace

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

Undefined

Solution steps

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

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

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

Evaluate

2 H (t) - 2 e^{- t}

at point

0 =

Undefined

= Undefined

y = 3 e^{4 x}

\int 6 x^{2} \cdot e^{x^{3}} d x

\int \frac{x ^{2} + x + 1}{( x + 1 ) ^{2} ( x + 2 )} d x

\frac{d}{d x} ((x cos (x)) (cos (x) + tan (x)))

\int_{0}^{1} 1 - x^{\frac{2}{7}} d x

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