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

解答

拉普拉斯变换

e^{- t} cos (2 t)

\frac{s + 1}{( s + 1 ) ^{2} + 4}

求解步骤

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

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

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

= \frac{( s + 1 )}{( s + 1 ) ^{2} + 4}

= \frac{s + 1}{( s + 1 ) ^{2} + 4}

\frac{\partial}{\partial x} (x^{\frac{1}{2}} y^{\frac{1}{2}} z)

\int_{0}^{6} 1 + x d x

x \to 3 + lim (\frac{x + 9}{x - 3})

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

f (x) = ln (3 x + 2)