laplacetransform t^2sin(2t)

解答

拉普拉斯变换

t^{2} sin (2 t)

- \frac{4 ( - 3 s ^{2} + 4 )}{( s ^{2} + 4 ) ^{3}}

求解步骤

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

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

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

\frac{d ^{2}}{d s ^{2}} (\frac{2}{s ^{2} + 4}) = - \frac{4 ( - 3 s ^{2} + 4 )}{( s ^{2} + 4 ) ^{3}}

= (- 1)^{2} (- \frac{4 ( - 3 s ^{2} + 4 )}{( s ^{2} + 4 ) ^{3}})

= - \frac{4 ( - 3 s ^{2} + 4 )}{( s ^{2} + 4 ) ^{3}}

x y^{^{'}} = y + 4 x^{2} sin (x), y (π) = 0

(1 + x^{2}) \frac{d y}{d x} = \frac{x ^{1}}{y}

\frac{d}{d x} (x^{4} - 13 x^{2} + 36)

a re a ∣ x ∣, 4 - x^{2}

a re a sin (x), cos (x), \frac{3 π}{4}, \frac{7 π}{6}