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

The answer to laplacetransform e^{2t}sin(t) is 1/((s-2)^2+1)

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laplacetransform e^{2t}sin(t)

laplace transform

e^{2 t} sin (t)

\frac{1}{( s - 2 ) ^{2} + 1}

Solution steps

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

Apply transform rule:

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

then

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

For

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

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

L {sin (t)} : \frac{1}{s ^{2} + 1}

= \frac{1}{( s - 2 ) ^{2} + 1}

\frac{d y}{d x}, y = \frac{3}{x}

x - y cos (\frac{y}{x}) d x + x cos (\frac{y}{x}) d y = 0

\frac{d}{d x} (\frac{x ^{2} - 4}{x - 3})

\frac{5}{x ^{2}}, at x = 1

\frac{t ^{4}}{24}

What is laplacetransform e^{2t}sin(t) ?
The answer to laplacetransform e^{2t}sin(t) is 1/((s-2)^2+1)