laplacetransform e^{-0.75t}cos(4t)

解答

拉普拉斯变换

e^{- 0.75 t} cos (4 t)

\frac{s + 0.75}{( s + 0.75 ) ^{2} + 16}

求解步骤

L {e^{- 0.75 t} cos (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^{- 0.75 t} cos (4 t) : f (t) = cos (4 t), a = - 0.75

= L {cos (4 t)} (s + 0.75)

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

= \frac{( s + 0.75 )}{( s + 0.75 ) ^{2} + 16}

= \frac{s + 0.75}{( s + 0.75 ) ^{2} + 16}

t an g e n t f (x) = \frac{1}{2} x^{2}, at x = 2

\int \frac{1 - x ^{2}}{1 - x} d x

\int \frac{t + 1}{3 t ^{2} + 6 t - 5} d t

\frac{\partial}{\partial y} (x e^{w})

x \to 2 - lim (\frac{3 x}{x - 2})