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

The answer to laplacetransform 2e^{-2t}sin(2t) is 4/((s+2)^2+4)

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

laplace transform

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

\frac{4}{( s + 2 ) ^{2} + 4}

Solution steps

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

Use the constant multiplication property of Laplace Transform:

For function

f (t)

and constant

a : L {a \cdot f (t)} = a \cdot L {f (t)}

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

L {e^{- 2 t} sin (2 t)} : \frac{2}{( s + 2 ) ^{2} + 4}

= 2 \cdot \frac{2}{( s + 2 ) ^{2} + 4}

Refine

2 \frac{2}{( s + 2 ) ^{2} + 4} : \frac{4}{( s + 2 ) ^{2} + 4}

= \frac{4}{( s + 2 ) ^{2} + 4}

n \to \infty lim (n^{2} - n)

\frac{1}{16 s ^{2} + 5 s + 20}

f (x) = \frac{1}{( 2 x ^{3} )}

x \to 5 lim (\frac{2}{( x - 5 ) ^{3}})

\frac{d}{d x} (2 - 4 x)

What is laplacetransform 2e^{-2t}sin(2t) ?
The answer to laplacetransform 2e^{-2t}sin(2t) is 4/((s+2)^2+4)