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

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

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

laplace transform

f (t) = 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}

\int \frac{cos ( 2 x )}{( 1 + sin ( 2 x ) ) ^{2}} d x

\frac{d}{d x} (12 x - x^{2})

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\frac{d}{d x} (\frac{3}{4} x^{\frac{4}{3}} - \frac{3}{8} x^{\frac{2}{3}} + 8)

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