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

The answer to laplacetransform e^{-2t}*sin(t) is 1/((s+2)^2+1)

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

laplace transform

e^{- 2 t} \cdot sin (t)

\frac{1}{( s + 2 ) ^{2} + 1}

Solution steps

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

Apply transform rule:

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

then

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

For

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

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

L {sin (t)} : \frac{1}{s ^{2} + 1}

= \frac{1}{( s + 2 ) ^{2} + 1}

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

x \to 0 lim ((tan (5 x))^{x})

\frac{1}{s ^{2} + 2 s + 2}

e^{c o s (x)}

\frac{\partial}{\partial x} (x^{2} - 3 x y + 4 yz + z^{3})

What is laplacetransform e^{-2t}*sin(t) ?
The answer to laplacetransform e^{-2t}*sin(t) is 1/((s+2)^2+1)