What is laplacetransform e^t*cos(t) ?

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

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laplacetransform e^t*cos(t)

laplace transform

e^{t} \cdot 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}

\frac{d}{d x} (\frac{2}{3} sin (\frac{x + 1}{x - 1}))

y^{^{''}} + 4 y^{^{'}} - 21 y = 0

\int csc (t) cot (t) d t

\int sin^{4} (x) d x

arcsin (sin (11 θ))

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