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

The answer to laplacetransform e^{-t}*sin(2t) is 2/((s+1)^2+4)

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

laplace transform

e^{- t} \cdot sin (2 t)

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

Solution steps

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

Apply transform rule:

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

then

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

For

e^{- t} sin (2 t) : f (t) = sin (2 t), a = - 1

= L {sin (2 t)} (s + 1)

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

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

\int \frac{x ^{2}}{x ^{2} + 6 x + 10} d x

x \to 0 + lim (x^{14} ln (x))

f (x) = - sin (2 x) \cdot 2

\frac{\partial}{\partial y} (cos (2 x - y))

x \to 1 lim (x^{2} - 4 x + 3)

What is laplacetransform e^{-t}*sin(2t) ?
The answer to laplacetransform e^{-t}*sin(2t) is 2/((s+1)^2+4)