What is laplacetransform e^{t/2}*cos(2t) ?

The answer to laplacetransform e^{t/2}*cos(2t) is (2s-1)/(2((s-1/2)^2+4))

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laplacetransform e^{t/2}*cos(2t)

laplace transform

e^{\frac{t}{2}} \cdot cos (2 t)

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

Solution steps

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

Apply transform rule:

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

then

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

For

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

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

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

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

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

\int 9 sin (ln (x)) d x

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f (x) = (x - 2)^{2}

(2 x + 3) y^{^{'}} = y + (2 x + 3)^{\frac{1}{2}}

What is laplacetransform e^{t/2}*cos(2t) ?
The answer to laplacetransform e^{t/2}*cos(2t) is (2s-1)/(2((s-1/2)^2+4))