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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse laplace

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

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

Solution steps

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

Expand

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

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

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

Simplify

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

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

\int \frac{x}{( x ^{2} + 1 ) ^{2}} d x

\frac{1}{x x}

\frac{d x}{d y} = (\frac{y + 1}{x}) 2

\frac{( x + 1 ) d y}{d x} + (x + 2) y = 2 x e^{- x}

\frac{\partial}{\partial x} (sin (x y) + x + y)

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