What is laplacetransform f(t)=2e^{-t}cos(4t)1(t) ?

The answer to laplacetransform f(t)=2e^{-t}cos(4t)1(t) is -(2(-s^2-2s+15))/((s^2+2s+17)^2)

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laplacetransform f(t)=2e^{-t}cos(4t)1(t)

laplace transform

f (t) = 2 e^{- t} cos (4 t) 1 (t)

- \frac{2 ( - s ^{2} - 2 s + 15 )}{( s ^{2} + 2 s + 17 ) ^{2}}

Solution steps

L {2 \cdot 1 \cdot e^{- t} (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 \cdot 1 L {e^{- t} cos (4 t) t}

L {e^{- t} cos (4 t) t} : - \frac{- s ^{2} - 2 s + 15}{( s ^{2} + 2 s + 17 ) ^{2}}

= 2 \cdot 1 \cdot (- \frac{- s ^{2} - 2 s + 15}{( s ^{2} + 2 s + 17 ) ^{2}})

Refine

21 (- \frac{- s ^{2} - 2 s + 15}{( s ^{2} + 2 s + 17 ) ^{2}}) : - \frac{2 ( - s ^{2} - 2 s + 15 )}{( s ^{2} + 2 s + 17 ) ^{2}}

= - \frac{2 ( - s ^{2} - 2 s + 15 )}{( s ^{2} + 2 s + 17 ) ^{2}}

\frac{d y}{d x} = (2 y + 1 cos (x))

\frac{d ^{2}}{d x ^{2}} (x^{3})

\int - \frac{6}{x ^{4}} d x

\int_{0}^{1} (x^{2} + y^{2} + z^{2}) d x

\int t^{2} e^{- 6 t} d t

What is laplacetransform f(t)=2e^{-t}cos(4t)1(t) ?
The answer to laplacetransform f(t)=2e^{-t}cos(4t)1(t) is -(2(-s^2-2s+15))/((s^2+2s+17)^2)