laplacetransform tcos(4t)

解答

拉普拉斯变换

t cos (4 t)

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

求解步骤

L {t cos (4 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 cos (4 t) : f (t) = cos (4 t), k = 1

= (- 1)^{1} \frac{d}{d s} (L {cos (4 t)})

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

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

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

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

y^{^{'}} = (0.5 + sin (t)) (\frac{y}{5})

t a y l or e^{- x}, x = 0

\int sec^{4} (\frac{x}{3}) d x

d er i v a t i v e y = x^{2} e^{x} - 2 x e^{x} + 5 e^{x}

\int_{- π}^{π} cos^{2} (x) d x