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

The answer to laplacetransform 2e^{-t}cos(4t) is (2(s+1))/((s+1)^2+16)

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

laplace transform

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

\frac{2 ( s + 1 )}{( s + 1 ) ^{2} + 16}

Solution steps

L {2 e^{- 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)}

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

L {e^{- t} cos (4 t)} : \frac{s + 1}{( s + 1 ) ^{2} + 16}

= 2 \cdot \frac{s + 1}{( s + 1 ) ^{2} + 16}

Refine

2 \frac{s + 1}{( s + 1 ) ^{2} + 16} : \frac{2 ( s + 1 )}{( s + 1 ) ^{2} + 16}

= \frac{2 ( s + 1 )}{( s + 1 ) ^{2} + 16}

\frac{d}{d x} (\frac{( 4 x + b )}{( c x + d )})

x y + \frac{d y}{d x} = 100 x

\frac{d y}{d x} + 8 y = e^{x} y^{- 8}

x \to 5 - lim (4 x - 5)

\int \frac{x}{( x - 1 ) ( x + 2 )} d x

What is laplacetransform 2e^{-t}cos(4t) ?
The answer to laplacetransform 2e^{-t}cos(4t) is (2(s+1))/((s+1)^2+16)