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

The answer to laplacetransform 5e^{-t}sin(2t) is (10)/((s+1)^2+4)

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

laplace transform

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

\frac{10}{( s + 1 ) ^{2} + 4}

Solution steps

L {5 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)}

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

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

= 5 \cdot \frac{2}{( s + 1 ) ^{2} + 4}

Refine

5 \frac{2}{( s + 1 ) ^{2} + 4} : \frac{10}{( s + 1 ) ^{2} + 4}

= \frac{10}{( s + 1 ) ^{2} + 4}

\frac{\partial}{\partial x} (sin (π (5 x - 4 y)))

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

(17 y^{2} - x y) d x + x^{2} d y = 0

\int \frac{e ^{3 x}}{5 + e ^{3 x}} d x

\frac{d}{d x} (arccsc (- 6 x^{2}))

What is laplacetransform 5e^{-t}sin(2t) ?
The answer to laplacetransform 5e^{-t}sin(2t) is (10)/((s+1)^2+4)