What is laplacetransform e^tcos(6t) ?

The answer to laplacetransform e^tcos(6t) is (s-1)/((s-1)^2+36)

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laplacetransform e^tcos(6t)

laplace transform

e^{t} cos (6 t)

\frac{s - 1}{( s - 1 ) ^{2} + 36}

Solution steps

L {e^{t} cos (6 t)}

Apply transform rule:

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

then

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

For

e^{t} cos (6 t) : f (t) = cos (6 t), a = 1

= L {cos (6 t)} (s - 1)

L {cos (6 t)} : \frac{s}{s ^{2} + 36}

= \frac{( s - 1 )}{( s - 1 ) ^{2} + 36}

= \frac{s - 1}{( s - 1 ) ^{2} + 36}

5 - x^{2}

f (x) = ln (x^{2} + 3)

y = (x + 1) (x + 2)

\int cos (200 t n) d t

\int sin (20 x) cos (17 x) d x

What is laplacetransform e^tcos(6t) ?
The answer to laplacetransform e^tcos(6t) is (s-1)/((s-1)^2+36)