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

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

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

laplace transform

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

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

Solution steps

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

Apply transform rule:

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

then

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

For

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

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

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

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

\int (- \frac{5}{x} + e^{- 2 x}) d x

\frac{\partial}{\partial x} (3 x y - 2 y)

\int \frac{1}{x ^{2} + 5 x} d x

\int - cot (x) csc (x) d x

1.4 x^{2} + 0.3 x + 0.25

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