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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse laplace

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

- \frac{1}{2} e^{- t} + \frac{1}{2} cos (t) + \frac{1}{2} sin (t)

Solution steps

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

Take the partial fraction of

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

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

Expand

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

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

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

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

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

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

\int (\frac{3 w ^{2}}{2} - \frac{2}{3 w ^{2}}) d w

\int x^{2} 2 + 9 x^{2} d x

y^{^{''}} - 7 y^{^{'}} + 12 y = 6 e t

\int \frac{x}{( sin ( x ) ) ^{2}} d x

\int x e^{x} + e^{x} d x

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