What is inverse oflaplace 1/(s^2+2s+5) ?

The answer to inverse oflaplace 1/(s^2+2s+5) is e^{-t} 1/2 sin(2t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 1/(s^2+2s+5)

inverse laplace

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

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

Solution steps

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

\frac{1}{s ^{2} + 2 s + 5} = \frac{1}{( s + 1 ) ^{2} + 4}

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{1}{( s + 1 ) ^{2} + 4} : a = - 1, F (s) = \frac{1}{s ^{2} + 4}

= e^{- t} L^{- 1} {\frac{1}{s ^{2} + 4}}

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

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

\frac{d}{d t} (t \cdot cos (t))

f (x) = \frac{5}{2 x}

\int (2 y - 3 y^{4} - y^{3}) \frac{1}{y} d y

y^{^{'}} + y + 3 cos (x) = 0

\frac{\partial}{\partial x} (x^{2} + 2 y sin (x))

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