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

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

inverse laplace

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

e^{- t} cos (t) + e^{- t} sin (t)

Solution steps

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

Expand

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

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

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

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

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

Refine

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

= e^{- t} cos (t) + e^{- t} sin (t)

x \to - 4 lim (7 - x)

\int_{- 2}^{2} x^{3} cos (\frac{x}{2}) d x

\int_{0}^{1} e^{- 0.2 x} d x

x \to \infty + lim (cos (x))

\int_{1}^{2} x ln (4 x) d x

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