laplacetransform te^{-3t}cos(5t)

解答

拉普拉斯变换

t e^{- 3 t} cos (5 t)

- \frac{- s ^{2} - 6 s + 16}{( s ^{2} + 6 s + 34 ) ^{2}}

求解步骤

L {e^{- 3 t} t cos (5 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 e^{- 3 t} cos (5 t) : f (t) = e^{- 3 t} cos (5 t), k = 1

= (- 1)^{1} \frac{d}{d s} (L {e^{- 3 t} cos (5 t)})

L {e^{- 3 t} cos (5 t)} : \frac{s + 3}{( s + 3 ) ^{2} + 25}

\frac{d}{d s} (\frac{s + 3}{( s + 3 ) ^{2} + 25}) = \frac{- s ^{2} - 6 s + 16}{( s ^{2} + 6 s + 34 ) ^{2}}

= (- 1)^{1} \frac{- s ^{2} - 6 s + 16}{( s ^{2} + 6 s + 34 ) ^{2}}

= - \frac{- s ^{2} - 6 s + 16}{( s ^{2} + 6 s + 34 ) ^{2}}

\frac{\partial}{\partial x} (ln (x + y + z))

\int x^{2} - 3 x + 2 d x

\int \frac{( x ^{2} - x + 5 )}{x} d x

y^{^{''}} - \frac{5}{x} y^{^{'}} + \frac{8}{x ^{2}} y = 0

7 t \frac{d y}{d t} - 3 y = t