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

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

inverse laplace

\frac{1}{( s + 1 ) ( s + 1 )}

e^{- t} t

Solution steps

L^{- 1} {\frac{1}{( s + 1 ) ( s + 1 )}}

Apply inverse transform rule:

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

then

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

For

\frac{1}{( s + 1 ) ( s + 1 )} : a = - 1, F (s) = \frac{1}{ss}

= e^{- t} L^{- 1} {\frac{1}{ss}}

L^{- 1} {\frac{1}{ss}} : t

= e^{- t} t

\frac{d y}{d x} = \frac{2 x ^{3}}{2 y + 1}

2 t y^{^{'}} - y = 0

\int x x^{2} + 25 d x

\int 5 x^{4} y^{3} - x^{2} y^{6} + 6 x^{5} - 4 y d x

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

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