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

The answer to inverse oflaplace (2s+1)/(s^2+2s+1)*1/s 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 (2s+1)/(s^2+2s+1)*1/s

inverse laplace

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

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

Solution steps

L^{- 1} {\frac{1}{s} \cdot \frac{2 s + 1}{s ^{2} + 2 s + 1}}

Expand

\frac{2 s + 1}{s ^{2} + 2 s + 1} \frac{1}{s} : \frac{2 s + 1}{s ^{3} + 2 s ^{2} + s}

= L^{- 1} {\frac{2 s + 1}{s ^{3} + 2 s ^{2} + s}}

Take the partial fraction of

\frac{2 s + 1}{s ^{3} + 2 s ^{2} + s} : \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}}}

Expand

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

\frac{1}{s ^{2} + 2 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

x = \frac{3}{y}, x = \frac{y}{9}, x = 9 y

\int 20

\frac{d}{d x} (sin (x + 3))

f (x) = ln (x - 2)

y^{^{'}} - \frac{2}{x} y = \frac{y ^{3}}{x ^{3}}

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