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

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

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

laplace transform

sin (5 t) e^{(- 2 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

sin (5 t) e^{- 2 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}

y = 2 e^{x} + x, (0, 2)

y = 6 x + 3

\int \frac{1}{16} d x

f (x) = \frac{48}{x ^{4}}

m = 1 \sum \infty 5^{m - 1}

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