What is laplacetransform e^{2t}cos(5t) ?

The answer to laplacetransform e^{2t}cos(5t) is (s-2)/((s-2)^2+25)

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laplacetransform e^{2t}cos(5t)

laplace transform

e^{2 t} cos (5 t)

\frac{s - 2}{( s - 2 ) ^{2} + 25}

Solution steps

L {e^{2 t} cos (5 t)}

Apply transform rule:

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

then

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

For

e^{2 t} cos (5 t) : f (t) = cos (5 t), a = 2

= L {cos (5 t)} (s - 2)

L {cos (5 t)} : \frac{s}{s ^{2} + 25}

= \frac{( s - 2 )}{( s - 2 ) ^{2} + 25}

= \frac{s - 2}{( s - 2 ) ^{2} + 25}

y = 3 x

x^{2} + y + 4 = 0, y = - 8

\frac{d}{d x} (cos^{22} (x) + 1)

x \to - 2 lim (6 x - x^{2})

y = 3 x^{3} - 5 x^{2} + 2 x + 3

What is laplacetransform e^{2t}cos(5t) ?
The answer to laplacetransform e^{2t}cos(5t) is (s-2)/((s-2)^2+25)