What is laplacetransform e^tsin(2t) ?

The answer to laplacetransform e^tsin(2t) is 2/((s-1)^2+4)

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laplacetransform e^tsin(2t)

laplace transform

e^{t} 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{1}{n ^{2}} cos (n x) d x

t \to \infty + lim (\frac{1}{4 t})

\frac{d}{d x} (a + b cos (x) + cs in (x))

\frac{d}{d x} (\frac{9}{1 - x})

x \to 5 + lim (\frac{1}{x - 5})

What is laplacetransform e^tsin(2t) ?
The answer to laplacetransform e^tsin(2t) is 2/((s-1)^2+4)