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

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

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

laplace transform

e^{2 t} cos (2 t)

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

Solution steps

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

Apply transform rule:

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

then

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

For

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

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

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

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

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

\int \frac{x}{2} + 1 d x

\int x^{6} - x^{4} d x

\frac{\partial}{\partial z} (x sin (z))

\int \frac{( t + 1 )}{t} d t

y^{^{''}} + k y (t) = 0

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