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

The answer to laplacetransform e^{-t}(cos(4t)-2sin(4t)) is (s+1)/((s+1)^2+16)-8/((s+1)^2+16)

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

laplace transform

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

\frac{s + 1}{( s + 1 ) ^{2} + 16} - \frac{8}{( s + 1 ) ^{2} + 16}

Solution steps

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

Apply transform rule:

L {f (t)} = F (s)

then

L {e^{a t} f (t)} = F (s - a)

For

e^{- t} (cos (4 t) - 2 sin (4 t)) : f (t) = (cos (4 t) - 2 sin (4 t)), a = - 1

= L {(cos (4 t) - 2 sin (4 t))} (s + 1)

L {(cos (4 t) - 2 sin (4 t))} : \frac{s}{s ^{2} + 16} - \frac{8}{s ^{2} + 16}

= \frac{( s + 1 )}{( s + 1 ) ^{2} + 16} - \frac{8}{( s + 1 ) ^{2} + 16}

= \frac{s + 1}{( s + 1 ) ^{2} + 16} - \frac{8}{( s + 1 ) ^{2} + 16}

\frac{d}{d t} (t^{2} e^{t})

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

\frac{d}{d x} ((2 + x)^{\frac{1}{3}} (1 - x)^{\frac{2}{3}})

\int \frac{9}{( x - 1 ) ( x + 8 )} d x

4 x^{4} + 3 x^{3} + 2 x^{2} + x + 1

What is laplacetransform e^{-t}(cos(4t)-2sin(4t)) ?
The answer to laplacetransform e^{-t}(cos(4t)-2sin(4t)) is (s+1)/((s+1)^2+16)-8/((s+1)^2+16)