laplacetransform t^2cos(5t)

解答

拉普拉斯变换

t^{2} cos (5 t)

- \frac{2 s ( - s ^{2} + 75 )}{( s ^{2} + 25 ) ^{3}}

求解步骤

L {t^{2} 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^{2} cos (5 t) : f (t) = cos (5 t), k = 2

= (- 1)^{2} \frac{d ^{2}}{d s ^{2}} (L {cos (5 t)})

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

\frac{d ^{2}}{d s ^{2}} (\frac{s}{s ^{2} + 25}) = - \frac{2 s ( - s ^{2} + 75 )}{( s ^{2} + 25 ) ^{3}}

= (- 1)^{2} (- \frac{2 s ( - s ^{2} + 75 )}{( s ^{2} + 25 ) ^{3}})

= - \frac{2 s ( - s ^{2} + 75 )}{( s ^{2} + 25 ) ^{3}}

d er i v a t i v e \frac{sin ( x )}{2 + cos ( x )}

t a y l or sin (5 x), x = 0

d er i v a t i v e \frac{1}{ln ( x + 1 )}

\frac{\partial}{\partial y} (5 x + y)

\frac{d}{d a} (sin^{4} (a) + cos^{4} (a))