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

The answer to inverse oflaplace (2s-1)/(s^2(s+1)^3) is 5\H(t)-t-5e^{-t}-e^{-t}*4t-(3e^{-t}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 (2s-1)/(s^2(s+1)^3)

inverse laplace

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

5 H (t) - t - 5 e^{- t} - e^{- t} \cdot 4 t - \frac{3 e ^{- t} t ^{2}}{2}

Solution steps

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

Take the partial fraction of

\frac{2 s - 1}{s ^{2} ( s + 1 ) ^{3}} : \frac{5}{s} - \frac{1}{s ^{2}} - \frac{5}{s + 1} - \frac{4}{( s + 1 ) ^{2}} - \frac{3}{( s + 1 ) ^{3}}

= L^{- 1} {\frac{5}{s} - \frac{1}{s ^{2}} - \frac{5}{s + 1} - \frac{4}{( s + 1 ) ^{2}} - \frac{3}{( s + 1 ) ^{3}}}

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{5}{s}} - L^{- 1} {\frac{1}{s ^{2}}} - L^{- 1} {\frac{5}{s + 1}} - L^{- 1} {\frac{4}{( s + 1 ) ^{2}}} - L^{- 1} {\frac{3}{( s + 1 ) ^{3}}}

L^{- 1} {\frac{5}{s}} : 5 H (t)

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

L^{- 1} {\frac{5}{s + 1}} : 5 e^{- t}

L^{- 1} {\frac{4}{( s + 1 ) ^{2}}} : e^{- t} \cdot 4 t

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

= 5 H (t) - t - 5 e^{- t} - e^{- t} \cdot 4 t - \frac{3 e ^{- t} t ^{2}}{2}

\int \frac{ln ( 4 x )}{x} d x

\frac{d}{d x} (x^{4} e^{5 x})

\int 2 sin (3 x) + 3 sin (2 x) d x

\frac{d}{d x} (ln (cosh (4 x)))

\int tan^{2} (2 y) d y

What is inverse oflaplace (2s-1)/(s^2(s+1)^3) ?
The answer to inverse oflaplace (2s-1)/(s^2(s+1)^3) is 5\H(t)-t-5e^{-t}-e^{-t}*4t-(3e^{-t}t^2)/2