What is inverse oflaplace 3/((s+1)^4) ?

The answer to inverse oflaplace 3/((s+1)^4) is (e^{-t}t^3)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 3/((s+1)^4)

inverse laplace

\frac{3}{( s + 1 ) ^{4}}

\frac{e ^{- t} t ^{3}}{2}

Solution steps

L^{- 1} {\frac{3}{( s + 1 ) ^{4}}}

Apply inverse transform rule:

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

then

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

For

\frac{3}{( s + 1 ) ^{4}} : a = - 1, F (s) = \frac{3}{s ^{4}}

= e^{- t} L^{- 1} {\frac{3}{s ^{4}}}

L^{- 1} {\frac{3}{s ^{4}}} : \frac{t ^{3}}{2}

= \frac{e ^{- t} t ^{3}}{2}

x \to 1 lim (4)

\frac{d x}{d t} + 5 t^{3} x^{5} + \frac{x}{t} = 0

\frac{d}{d x} ((2 - x^{2}) (4 x^{2} + 9 x)^{4})

\frac{d}{d x} (\frac{5 ^{x}}{x ^{5}})

\int (x^{4} + x^{2} + x) d x

What is inverse oflaplace 3/((s+1)^4) ?
The answer to inverse oflaplace 3/((s+1)^4) is (e^{-t}t^3)/2