What is laplacetransform e^{-t}sin(t) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

laplacetransform e^{-t}sin(t)

laplace transform

e^{- t} sin (t)

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

Solution steps

L {e^{- t} sin (t)}

Apply transform rule:

L {f (t)} = F (s)

then

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

For

e^{- t} sin (t) : f (t) = sin (t), a = - 1

= L {sin (t)} (s + 1)

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

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

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