What is laplacetransform e^{-2t}sin(t) ?

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

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

laplace transform

e^{- 2 t} sin (t)

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

Solution steps

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

Apply transform rule:

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

then

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

For

e^{- 2 t} sin (t) : f (t) = sin (t), a = - 2

= L {sin (t)} (s + 2)

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

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

x \to a lim (\frac{3}{0})

\frac{20}{s ( s + 10 )}

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

n = 0 \sum \infty \frac{1}{( 1 + n ^{2} )}

t^{- 2}

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