What is inverse oflaplace 1/((s+1)^3) ?

The answer to inverse oflaplace 1/((s+1)^3) is (e^{-t}t^2)/2

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)^3)

inverse laplace

\frac{1}{( s + 1 ) ^{3}}

\frac{e ^{- t} t ^{2}}{2}

Solution steps

L^{- 1} {\frac{1}{( s + 1 ) ^{3}}}

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 ) ^{3}} : a = - 1, F (s) = \frac{1}{s ^{3}}

= e^{- t} L^{- 1} {\frac{1}{s ^{3}}}

L^{- 1} {\frac{1}{s ^{3}}} : \frac{t ^{2}}{2}

= \frac{e ^{- t} t ^{2}}{2}

n = 1 \sum \infty \frac{2}{n ^{\frac{3}{2}}}

x y^{2} y^{^{'}} = x + 1

\int \frac{9 - 4 x}{x} d x

y = 9 e^{x} + x, (0, 9)

\frac{s + 5}{s ( s + 2 ) ( s + 3 )}

What is inverse oflaplace 1/((s+1)^3) ?
The answer to inverse oflaplace 1/((s+1)^3) is (e^{-t}t^2)/2