What is laplacetransform e^tcos(t) ?

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

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laplacetransform e^tcos(t)

laplace transform

e^{t} cos (t)

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

Solution steps

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

Apply transform rule:

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

then

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

For

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

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

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

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

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

\int sin (\frac{nπ x}{π}) d x

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\int 5 yz d x

(\frac{x ^{2}}{x - 5})^{^{'}}

x^{7} (x - 10)^{3}

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