What is inverse oflaplace s/(s^2+2s+3) ?

The answer to inverse oflaplace s/(s^2+2s+3) is e^{-t}cos(sqrt(2)t)-1/(sqrt(2))e^{-t}sin(sqrt(2)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^2+2s+3)

inverse laplace

\frac{s}{s ^{2} + 2 s + 3}

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

Solution steps

L^{- 1} {\frac{s}{s ^{2} + 2 s + 3}}

Expand

\frac{s}{s ^{2} + 2 s + 3} : \frac{s + 1}{( s + 1 ) ^{2} + 2} - 1 \cdot \frac{1}{( s + 1 ) ^{2} + 2}

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

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

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

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

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

Refine

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

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

f (x) = \frac{2 x ^{3} - 3 x + 1}{x}

y = x^{2} - 2

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

f (x) = x + 7 x

\frac{d}{d x} (- x^{3} + 3 x)

What is inverse oflaplace s/(s^2+2s+3) ?
The answer to inverse oflaplace s/(s^2+2s+3) is e^{-t}cos(sqrt(2)t)-1/(sqrt(2))e^{-t}sin(sqrt(2)t)