laplacetransform e^{-2t}cos(5t)

解答

拉普拉斯变换

e^{- 2 t} cos (5 t)

\frac{s + 2}{( s + 2 ) ^{2} + 25}

求解步骤

L {e^{- 2 t} cos (5 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^{- 2 t} cos (5 t) : f (t) = cos (5 t), a = - 2

= L {cos (5 t)} (s + 2)

L {cos (5 t)} : \frac{s}{s ^{2} + 25}

= \frac{( s + 2 )}{( s + 2 ) ^{2} + 25}

= \frac{s + 2}{( s + 2 ) ^{2} + 25}

\frac{d}{d x} (4 \frac{5}{4} x)

\int \frac{1}{x ^{2} - 2 x + 26} d x

x^{^{''}} - a x = 0

x \to 0 lim (x^{2} cos (\frac{8}{x}))

d er i v a t i v e y = \frac{3 x - 2}{x}