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

The answer to laplacetransform e^{2t}sin(4t) is 4/((s-2)^2+16)

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

laplace transform

e^{2 t} sin (4 t)

\frac{4}{( s - 2 ) ^{2} + 16}

Solution steps

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

Apply transform rule:

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

then

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

For

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

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

L {sin (4 t)} : \frac{4}{s ^{2} + 16}

= \frac{4}{( s - 2 ) ^{2} + 16}

(x^{3} - 2)^{2}

\frac{d}{d x} (x - 6 x)

x \to 1 lim (x^{3})

x \to + 3 lim (\frac{5}{x ^{2} - 9})

\frac{d y}{d t} = 20 - \frac{y}{230 + 16 t}

What is laplacetransform e^{2t}sin(4t) ?
The answer to laplacetransform e^{2t}sin(4t) is 4/((s-2)^2+16)