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

解答

拉普拉斯变换

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

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

求解步骤

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

应用变换法则: if

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

则

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

对于

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}

n = 3 \sum \infty \frac{1}{n ^{2}}

d er i v a t i v e F (x) = 5 x^{5} (x^{3} - 9 x)

x \to \infty lim (\frac{2 x + 1}{2})

\frac{d}{d x} ((2 + (x^{2} + 1)^{4})^{3})

\int 2 x sin (4 x) d x