What is inverse oflaplace (e^{-s})/(s*(s+5)^2) ?

The answer to inverse oflaplace (e^{-s})/(s*(s+5)^2) is (\H(t-1)(1-5e^{-5(t-1)}t+4e^{-5(t-1)}))/(25)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace (e^{-s})/(s*(s+5)^2)

inverse laplace

\frac{e ^{- s}}{s \cdot ( s + 5 ) ^{2}}

\frac{H ( t - 1 ) ( 1 - 5 e ^{- 5 (t - 1)} t + 4 e ^{- 5 (t - 1)} )}{25}

Solution steps

L^{- 1} {\frac{e ^{- s}}{s ( s + 5 ) ^{2}}}

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 - 1) L^{- 1} {\frac{1}{s ( s + 5 ) ^{2}}} (t - 1)

L^{- 1} {\frac{1}{s ( s + 5 ) ^{2}}} : \frac{1}{25} H (t) - \frac{1}{25} e^{- 5 t} - \frac{e ^{- 5 t} t}{5}

= H (t - 1) (\frac{1}{25} H (t - 1) - \frac{1}{25} e^{- 5 (t - 1)} - \frac{e ^{- 5 (t - 1)} ( t - 1 )}{5})

H (t - 1) = {01 t < 1 t \geq 1

= {0 \frac{1}{25} - \frac{1}{25} e^{- 5 (t - 1)} - \frac{e ^{- 5 (t - 1)} ( t - 1 )}{5} t < 1 t \geq 1

= H (t - 1) (\frac{1}{25} - \frac{1}{25} e^{- 5 (t - 1)} - \frac{e ^{- 5 (t - 1)} ( t - 1 )}{5})

Simplify

H (t - 1) (\frac{1}{25} - \frac{1}{25} e^{- 5 (t - 1)} - \frac{e ^{- 5 (t - 1)} ( t - 1 )}{5}) : \frac{H ( t - 1 ) ( 1 - 5 e ^{- 5 (t - 1)} t + 4 e ^{- 5 (t - 1)} )}{25}

= \frac{H ( t - 1 ) ( 1 - 5 e ^{- 5 (t - 1)} t + 4 e ^{- 5 (t - 1)} )}{25}

\frac{d}{d x} (\frac{ln ( 3 x )}{3 x})

\int \frac{16}{( x + 2 ) ( x ^{2} - 4 )} d x

\int e^{6 x} sin (6 x) d x

\frac{d y}{d x} - y^{2} = 0

(\frac{x ^{2} + 3}{x ^{2} - 3})^{5}

What is inverse oflaplace (e^{-s})/(s*(s+5)^2) ?
The answer to inverse oflaplace (e^{-s})/(s*(s+5)^2) is (\H(t-1)(1-5e^{-5(t-1)}t+4e^{-5(t-1)}))/(25)