laplacetransform t^2cos(kt)

解答

拉普拉斯变换

t^{2} cos (k t)

- \frac{2 s ( - s ^{2} + 3 k ^{2} )}{( s ^{2} + k ^{2} ) ^{3}}

求解步骤

L {t^{2} cos (k t)}

使用拉普拉斯变换表:

L {t^{k} f (t)} = (- 1)^{k} \frac{d ^{k}}{d s ^{k}} (L {f (t)})

对于

t^{2} cos (k t) : f (t) = cos (k t), k = 2

= (- 1)^{2} \frac{d ^{2}}{d s ^{2}} (L {cos (k t)})

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

\frac{d ^{2}}{d s ^{2}} (\frac{s}{s ^{2} + k ^{2}}) = - \frac{2 s ( - s ^{2} + 3 k ^{2} )}{( s ^{2} + k ^{2} ) ^{3}}

= (- 1)^{2} (- \frac{2 s ( - s ^{2} + 3 k ^{2} )}{( s ^{2} + k ^{2} ) ^{3}})

= - \frac{2 s ( - s ^{2} + 3 k ^{2} )}{( s ^{2} + k ^{2} ) ^{3}}

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

\int x 5 - 2 x d x

\int_{0}^{7} x e^{- 4 x} d x

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

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