What is laplacetransform e^{-5t}sin(2t) ?

The answer to laplacetransform e^{-5t}sin(2t) is 2/((s+5)^2+4)

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laplacetransform e^{-5t}sin(2t)

laplace transform

e^{- 5 t} sin (2 t)

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

Solution steps

L {e^{- 5 t} sin (2 t)}

Apply transform rule:

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

then

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

For

e^{- 5 t} sin (2 t) : f (t) = sin (2 t), a = - 5

= L {sin (2 t)} (s + 5)

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

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

\int - x^{2} + 4 x d x

(2 x y - sec^{2} (x)) d x + (x^{2} + 2 y) d y = 0

\int 2^{s i n^{2} (\frac{1}{3}) x} sin (\frac{2}{3}) x d x

What is laplacetransform e^{-5t}sin(2t) ?
The answer to laplacetransform e^{-5t}sin(2t) is 2/((s+5)^2+4)