laplacetransform t^2cos(3t)

解答

拉普拉斯变换

t^{2} cos (3 t)

- \frac{2 s ( - s ^{2} + 27 )}{( s ^{2} + 9 ) ^{3}}

求解步骤

L {t^{2} cos (3 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 (3 t) : f (t) = cos (3 t), k = 2

= (- 1)^{2} \frac{d ^{2}}{d s ^{2}} (L {cos (3 t)})

L {cos (3 t)} : \frac{s}{s ^{2} + 9}

\frac{d ^{2}}{d s ^{2}} (\frac{s}{s ^{2} + 9}) = - \frac{2 s ( - s ^{2} + 27 )}{( s ^{2} + 9 ) ^{3}}

= (- 1)^{2} (- \frac{2 s ( - s ^{2} + 27 )}{( s ^{2} + 9 ) ^{3}})

= - \frac{2 s ( - s ^{2} + 27 )}{( s ^{2} + 9 ) ^{3}}

x \to 13 lim (x^{4} - 10 x^{2} - 20)

x \to \infty lim (x e^{(\frac{1}{x})})

in v erse l a pl a ce \frac{s ^{2} + 4}{( s ^{2} - 4 ) ^{2}}

5 x - 8 y x^{2} + 1 \frac{d y}{d x} = 0

s im pl i f y 2 ln (\frac{3 x ^{2}}{2})