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

The answer to laplacetransform 2e^{-t}sin(2t) is 4/((s+1)^2+4)

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

laplace transform

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

\frac{4}{( s + 1 ) ^{2} + 4}

Solution steps

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

L {e^{- t} sin (2 t)} : \frac{2}{( s + 1 ) ^{2} + 4}

= 2 \cdot \frac{2}{( s + 1 ) ^{2} + 4}

Refine

2 \frac{2}{( s + 1 ) ^{2} + 4} : \frac{4}{( s + 1 ) ^{2} + 4}

= \frac{4}{( s + 1 ) ^{2} + 4}

(1 + t e^{- t})^{3}

\int \frac{2 x}{x - 1} d x

t \frac{d y}{d t} + 34 y = \frac{1}{t}

f (x) = x - 1, at x = 5

\int (5 - 6 x) \cdot sin (4 x) d x

What is laplacetransform 2e^{-t}sin(2t) ?
The answer to laplacetransform 2e^{-t}sin(2t) is 4/((s+1)^2+4)