What is inverse oflaplace 8/((2s+1)^3) ?

The answer to inverse oflaplace 8/((2s+1)^3) is (e^{-t/2}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 8/((2s+1)^3)

inverse laplace

\frac{8}{( 2 s + 1 ) ^{3}}

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

Solution steps

L^{- 1} {\frac{8}{( 2 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{8}{( 2 s + 1 ) ^{3}} : a = - \frac{1}{2}, F (s) = \frac{8}{( 2 s ) ^{3}}

= e^{- \frac{1}{2} t} L^{- 1} {\frac{8}{( 2 s ) ^{3}}}

L^{- 1} {\frac{8}{( 2 s ) ^{3}}} : \frac{t ^{2}}{2}

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

\frac{\partial}{\partial x} (2 x + 3 y - 4)

f (x) \frac{2}{x}

\frac{x ^{2} - 2 x - 3}{( x - 1 ) ^{2}}

144 y^{^{'''}} + 168 y^{^{''}} + 49 y^{^{'}} = 0

\frac{d}{d x} (\frac{3 x}{1 + x ^{2}})

What is inverse oflaplace 8/((2s+1)^3) ?
The answer to inverse oflaplace 8/((2s+1)^3) is (e^{-t/2}t^2)/2