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

The answer to inverse oflaplace s/(s^2-2s+3) is e^tcos(sqrt(2)t)+1/(sqrt(2))e^tsin(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)

\frac{\partial}{\partial x} (4 x^{2})

\int 4 + 9 x^{2} d x

\int sin (3 x) cos (4 x) sin (2 x) d x

\frac{d}{d x} (\frac{x \cdot cos ( x ) + π}{x - π})

\frac{r}{r ^{2} + 7}

What is inverse oflaplace s/(s^2-2s+3) ?
The answer to inverse oflaplace s/(s^2-2s+3) is e^tcos(sqrt(2)t)+1/(sqrt(2))e^tsin(sqrt(2)t)