What is laplacetransform e^tcos(2t) ?

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

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laplacetransform e^tcos(2t)

laplace transform

e^{t} cos (2 t)

\frac{s - 1}{( s - 1 ) ^{2} + 4}

Solution steps

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

Apply transform rule:

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

then

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

For

e^{t} cos (2 t) : f (t) = cos (2 t), a = 1

= L {cos (2 t)} (s - 1)

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

= \frac{( s - 1 )}{( s - 1 ) ^{2} + 4}

= \frac{s - 1}{( s - 1 ) ^{2} + 4}

\frac{d}{d x} (e^{0 x})

f (x) = \frac{π x}{2}

\frac{d}{d x} (64 x - \frac{x ^{3}}{3})

\frac{d}{d x} (ln (6 x^{2} + 3 x))

x \to 1 lim ((x - 2)^{2})

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