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

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

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

laplace transform

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

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

Solution steps

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

f (x) = 2 x^{3} + 3 x^{2} + 4 x + 2, at x = - 1

y^{^{''}} + 3 y^{^{'}} = 9 x

\frac{d}{d x} (e^{(- 5 x^{2} - 2 x^{2})})

\frac{d}{d y} (\frac{y - 2 ( u y )}{y})

y^{^{''}} + 2 y = 18

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