What is inverse oflaplace (e^{-4s})/(s^3) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace (e^{-4s})/(s^3)

inverse laplace

\frac{e ^{- 4 s}}{s ^{3}}

\frac{H ( t - 4 ) ( t - 4 ) ^{2}}{2}

Solution steps

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

Apply inverse transform rule:

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

then

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

Where

H (t)

is Heaviside step function

= H (t - 4) L^{- 1} {\frac{1}{s ^{3}}} (t - 4)

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

= H (t - 4) \frac{( t - 4 ) ^{2}}{2}

= \frac{H ( t - 4 ) ( t - 4 ) ^{2}}{2}

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

\int tan^{3} (d) x d x

\frac{d}{d θ} (r (θ) cos (θ))

3 x + y - 2 + y^{^{'}} (x - 1) = 0

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

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