What is laplacetransform e^{4t}sin(2t) ?

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

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laplacetransform e^{4t}sin(2t)

laplace transform

e^{4 t} sin (2 t)

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

Solution steps

L {e^{4 t} sin (2 t)}

Apply transform rule:

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

then

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

For

e^{4 t} sin (2 t) : f (t) = sin (2 t), a = 4

= L {sin (2 t)} (s - 4)

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

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

\frac{d}{d y} (\frac{y}{t ^{2} + y ^{2}})

\frac{d r}{d t} + 2 t r - r = 0, r (0) = 6

\frac{\partial}{\partial y} (sin (3 x - 4 y))

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

x \to 2 lim (\frac{- 2}{x ( x - 2 )})

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