What is inverse oflaplace s/(s^2-2s+5) ?

The answer to inverse oflaplace s/(s^2-2s+5) is e^tcos(2t)+1/2 e^tsin(2t)

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+5)

inverse laplace

\frac{s}{s ^{2} - 2 s + 5}

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

Solution steps

L^{- 1} {\frac{s}{s ^{2} - 2 s + 5}}

Expand

\frac{s}{s ^{2} - 2 s + 5} : \frac{s - 1}{( s - 1 ) ^{2} + 4} + 1 \cdot \frac{1}{( s - 1 ) ^{2} + 4}

= L^{- 1} {\frac{s - 1}{( s - 1 ) ^{2} + 4} + 1 \cdot \frac{1}{( s - 1 ) ^{2} + 4}}

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} + 4}} + 1 \cdot L^{- 1} {\frac{1}{( s - 1 ) ^{2} + 4}}

L^{- 1} {\frac{s - 1}{( s - 1 ) ^{2} + 4}} : e^{t} cos (2 t)

L^{- 1} {\frac{1}{( s - 1 ) ^{2} + 4}} : 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)

(8 - 2 x^{3})^{3}

\int \frac{1}{x 9 + x ^{2}} d x

\frac{\partial}{\partial x} (x y - ln (x^{2} - y^{2}))

5 x^{2} y^{^{'}} = y^{^{'}} + 6 x e^{- y}

\frac{\partial}{\partial x} ((x^{2} + y^{2} + z^{2})^{- \frac{1}{4}})

What is inverse oflaplace s/(s^2-2s+5) ?
The answer to inverse oflaplace s/(s^2-2s+5) is e^tcos(2t)+1/2 e^tsin(2t)