What is laplacetransform 3e^{-2t}cos(4t) ?

The answer to laplacetransform 3e^{-2t}cos(4t) is (3(s+2))/((s+2)^2+16)

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laplacetransform 3e^{-2t}cos(4t)

laplace transform

3 e^{- 2 t} cos (4 t)

\frac{3 ( s + 2 )}{( s + 2 ) ^{2} + 16}

Solution steps

L {3 e^{- 2 t} cos (4 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)}

= 3 L {e^{- 2 t} cos (4 t)}

L {e^{- 2 t} cos (4 t)} : \frac{s + 2}{( s + 2 ) ^{2} + 16}

= 3 \cdot \frac{s + 2}{( s + 2 ) ^{2} + 16}

Refine

3 \frac{s + 2}{( s + 2 ) ^{2} + 16} : \frac{3 ( s + 2 )}{( s + 2 ) ^{2} + 16}

= \frac{3 ( s + 2 )}{( s + 2 ) ^{2} + 16}

\int e^{x + y + z} d x

\int_{0}^{1} (7 e^{x} - 7 x e^{x^{2}}) d x

f (x) = 3 x ln (19 x)

\int 3 x^{2} cos (2 x) d x

\frac{d y}{d x} - 2 x y = 4 x

What is laplacetransform 3e^{-2t}cos(4t) ?
The answer to laplacetransform 3e^{-2t}cos(4t) is (3(s+2))/((s+2)^2+16)