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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

inverse laplace

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

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

Solution steps

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

Expand

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

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

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

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

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

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