What is inverse oflaplace (2s+4)/((s+1)(s^2+4)) ?

The answer to inverse oflaplace (2s+4)/((s+1)(s^2+4)) is 2/5 e^{-t}-2/5 cos(2t)+(6sin(2t))/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse oflaplace (2s+4)/((s+1)(s^2+4))

inverse laplace

\frac{2 s + 4}{( s + 1 ) ( s ^{2} + 4 )}

\frac{2}{5} e^{- t} - \frac{2}{5} cos (2 t) + \frac{6 sin ( 2 t )}{5}

Solution steps

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

Take the partial fraction of

\frac{2 s + 4}{( s + 1 ) ( s ^{2} + 4 )} : \frac{2}{5 ( s + 1 )} + \frac{- 2 s + 12}{5 ( s ^{2} + 4 )}

= L^{- 1} {\frac{2}{5 ( s + 1 )} + \frac{- 2 s + 12}{5 ( s ^{2} + 4 )}}

Expand

= L^{- 1} {- \frac{2 s}{5 ( s ^{2} + 4 )} + \frac{2}{5 ( s + 1 )} + \frac{12}{5 ( s ^{2} + 4 )}}

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)}

= \frac{2}{5} L^{- 1} {\frac{1}{s + 1}} - \frac{2}{5} L^{- 1} {\frac{s}{s ^{2} + 4}} + \frac{12}{5} L^{- 1} {\frac{1}{s ^{2} + 4}}

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

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

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

= \frac{2}{5} e^{- t} - \frac{2}{5} cos (2 t) + \frac{12}{5} \cdot \frac{1}{2} sin (2 t)

Refine

\frac{2}{5} e^{- t} - \frac{2}{5} cos (2 t) + \frac{12}{5} \frac{1}{2} sin (2 t) : \frac{2}{5} e^{- t} - \frac{2}{5} cos (2 t) + \frac{6 sin ( 2 t )}{5}

= \frac{2}{5} e^{- t} - \frac{2}{5} cos (2 t) + \frac{6 sin ( 2 t )}{5}

What is inverse oflaplace (2s+4)/((s+1)(s^2+4)) ?
The answer to inverse oflaplace (2s+4)/((s+1)(s^2+4)) is 2/5 e^{-t}-2/5 cos(2t)+(6sin(2t))/5