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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

\frac{1}{( 1 - s ) ( 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)

y^{^{'}} (x) = \frac{9 x ^{2}}{9 y ^{2} - 7}, y (1) = 0

\int \frac{x + x ^{2}}{x ^{3}} d x

\int_{- 1}^{3} (x + 1)^{3} d x

(x^{4} + 2 y) d x - x d y = 0

0.2 x^{3} \cdot 3^{x}

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