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

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

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laplacetransform e^{-2t}sin(5t)

laplace transform

e^{- 2 t} sin (5 t)

\frac{5}{( s + 2 ) ^{2} + 25}

Solution steps

L {e^{- 2 t} sin (5 t)}

Apply transform rule:

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

then

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

For

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

= L {sin (5 t)} (s + 2)

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

= \frac{5}{( s + 2 ) ^{2} + 25}

6 e^{x^{- 5}}

f (x) = \frac{1}{3 + 2 x}, (1, \frac{1}{5})

\frac{d}{d x} (\frac{c x + b}{4 x + d})

\frac{d ^{2}}{d x ^{2}} (cos (x^{2}))

\int - x^{5} sin (x) d x

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