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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 1/(s*(s+1)^2)

inverse laplace

\frac{1}{s \cdot ( s + 1 ) ^{2}}

H (t) - e^{- t} - e^{- t} t

Solution steps

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

Take the partial fraction of

\frac{1}{s ( s + 1 ) ^{2}} : \frac{1}{s} - \frac{1}{s + 1} - \frac{1}{( s + 1 ) ^{2}}

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

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

L^{- 1} {\frac{1}{s}} : H (t)

L^{- 1} {\frac{1}{s + 1}} : e^{- t}

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

= H (t) - e^{- t} - e^{- t} t

9^{x}

\int \frac{sin ^{2} ( θ )}{cos ( θ )} d θ

x \to 5 lim (x^{2} + 24)

\int sin (7 x) \cdot cos (3 x) d x

\frac{d}{d x} (∣ x - 5 ∣)

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