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}

\int \frac{1}{x ^{2} - 2 x + 26} d x

x^{^{''}} - a x = 0

x \to 0 lim (x^{2} cos (\frac{8}{x}))

y = \frac{3 x - 2}{x}

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